Suppose I have an array of integers:
int[] A = { 10, 3, 6, 8, 9, 4, 3 };
My goal is to find the largest difference between A[Q] and A[P] such that Q > P.
For example, if P = 2 and Q = 3, then
diff = A[Q] - A[P]
diff = 8 - 6
diff = 2
If P = 1 and Q = 4
diff = A[Q] - A[P]
diff = 9 - 3
diff = 6
Since 6 is the largest number between all the difference, that is the answer.
My solution is as follows (in C#) but it is inefficient.
public int solution(int[] A) {
int N = A.Length;
if (N < 1) return 0;
int difference;
int largest = 0;
for (int p = 0; p < N; p++)
{
for (int q = p + 1; q < N; q++)
{
difference = A[q] - A[p];
if (difference > largest)
{
largest = difference;
}
}
}
return largest;
}
How can I improve this so it will run at O(N)? Thanks!
Simply getting the max and min wont work. Minuend (Q) should come after the Subtrahend (P).
This question is based on the "Max-profit" problem in codility (http://codility.com/train/). My solution only scored 66%. It requires O(N) for a score of 100%.
The following code runs in O(n) and should conform to the specification (preliminary tests on codility were successful):
public int solution(int[] A)
{
int N = A.Length;
if (N < 1) return 0;
int max = 0;
int result = 0;
for(int i = N-1; i >= 0; --i)
{
if(A[i] > max)
max = A[i];
var tmpResult = max - A[i];
if(tmpResult > result)
result = tmpResult;
}
return result;
}
Update:
I submitted it as solution and it scores 100%.
Update 02/26/16:
The original task description on codility stated that "each element of array A is an integer within the range [0..1,000,000,000]."
If negative values would have been allowed as well, the code above wouldn't return the correct value. This could be fixed easily by changing the declaration of max to int max = int.MinValue;
Here is the O(n) Java implementation
public static int largestDifference(int[] data) {
int minElement=data[0], maxDifference=0;
for (int i = 1; i < data.length; i++) {
minElement = Math.min(minElement, data[i]);
maxDifference = Math.max(maxDifference, data[i] - minElement);
}
return maxDifference;
}
After some attempts, I end up with this:
int iMax = N - 1;
int min = int.MaxValue, max = int.MinValue;
for (int i = 0; i < iMax; i++) {
if (min > A[i]) min = A[i];
if (max < A[N - i - 1]){
iMax = N - i - 1;
max = A[iMax];
}
}
int largestDiff = max - min;
NOTE: I have just tested it with some cases. Please if you find any case in which it doesn't work, let me know in the comment. I'll try to improve it or remove the answer. Thanks!
int FirstIndex = -1;
int SecondIndex = -1;
int diff = 0;
for (int i = A.Length-1; i >=0; i--)
{
int FirstNo = A[i];
int tempDiff = 0;
for (int j = 0; j <i ; j++)
{
int SecondNo = A[j];
tempDiff = FirstNo - SecondNo;
if (tempDiff > diff)
{
diff = tempDiff;
FirstIndex = i;
SecondIndex = j;
}
}
}
MessageBox.Show("Diff: " + diff + " FirstIndex: " + (FirstIndex+1) + " SecondIndex: " + (SecondIndex+1));
PHP Solution
<?php
$a = [0,5,0,5,0];
$max_diff = -1;
$min_value = $a[0];
for($i = 0;$i<count($a)-1;$i++){
if($a[$i+1] > $a[$i]){
$diff = $a[$i+1] - $min_value;
if($diff > $max_diff){
$max_diff = $diff;
}
} else {
$min_value = $a[$i+1];
}
}
echo $max_diff;
?>
We can do it in a much simpler way by calculating biggest and smallest element of the array. I know that you're also looking for time complexity. But for anyone looking to understand and solve this problem in a simple and easy to understand way, then here is my code:
#include<stdio.h>
#define N 6
int main()
{
int num[N], i, big, small, pos = 0;
printf("Enter %d integer numbers\n", N);
for(i = 0; i < N; i++)
scanf("%d", &num[i]);
big = small = num[0];
for(i = 1; i < N; i++)
{
if(num[i] > big)
{
big = num[i];
pos = i;
}
}
for(i = 1; i < pos; i++)
{
if(num[i] < small)
small = num[i];
}
printf("The largest difference is %d, ", (big - small));
printf("and its between %d and %d.\n", big, small);
return 0;
}
Output:
Enter 6 integer numbers
7
9
5
6
13
2
The largest difference is 8, and its between 13 and 5.
Source: C Program To Find Largest Difference Between Two Elements of Array
C++ solution for MaxProfit of codility test task giving 100/100 https://app.codility.com/programmers/lessons/9-maximum_slice_problem/max_profit/
int Max(vector<int> &A)
{
if (A.size() == 1 || A.size() == 0)
return 0;
int min_price = A[0];
int max_val = 0;
for (int i = 1; i < A.size(); i++)
{
max_val = std::max(max_val, A[i] - min_price);
min_price = std::min(min_price, A[i]);
}
return max_val;
}
My 100% JavaScript solution with O(N) time complexity:
function solution(A) {
// each element of array A is an integer within the range [0..200,000]
let min = 200000;
// The function should return 0 if it was impossible to gain any profit.
let maxDiff = 0;
for (const a of A) {
min = Math.min(min, a);
// find the maximum positive difference (profit) between current global minimum and current value of a
maxDiff = Math.max(maxDiff, a - min);
}
return maxDiff;
}
function solution(A) {
var n = A.length;
var min = Infinity, max = -Infinity, maxNet=0;
// find smallest and largest in the array following each other
for(let i = 0; i < n; i++){
if(A[i]<min) { // if you are updating the min you cannot consider the old max
min = A[i];
max = -Infinity;
} else if(A[i]> max){
max = A[i];
}
if(max!=-Infinity && max-min>maxNet) maxNet = max-min;
}
return maxNet;
}
PHP solution for MaxProfit of codility test task giving 100/100 found at http://www.rationalplanet.com/php-related/maxprofit-demo-task-at-codility-com.html
function solution($A) {
$cnt = count($A);
if($cnt == 1 || $cnt == 0){
return 0;
}
$max_so_far = 0;
$max_ending_here = 0;
$min_price = $A[0];
for($i = 1; $i < $cnt; $i++){
$max_ending_here = max(0, $A[$i] - $min_price);
$min_price = min($min_price, $A[$i]);
$max_so_far = max($max_ending_here, $max_so_far);
}
return $max_so_far;
}
100% score JavaScript solution.
function solution(A) {
if (A.length < 2)
return 0;
// Init min price and max profit
var minPrice = A[0];
var maxProfit = 0;
for (var i = 1; i < A.length; i++) {
var profit = A[i] - minPrice;
maxProfit = Math.max(maxProfit, profit);
minPrice = Math.min(minPrice, A[i]);
}
return maxProfit;
}
Python solution
def max_diff_two(arr):
#keep tab of current diff and min value
min_value = arr[0]
#begin with something
maximum = arr[1] - arr[0]
new_min = min_value
for i,value in enumerate(arr):
if i == 0:
continue
if value < min_value and value < new_min:
new_min = value
current_maximum = value - min_value
new_maximum = value - new_min
if new_maximum > current_maximum:
if new_maximum > maximum:
maximum = new_maximum
min = new_min
else:
if current_maximum > maximum:
maximum = current_maximum
return maximum
100% for Javascript solution using a more elegant functional approach.
function solution(A) {
var result = A.reverse().reduce(function (prev, val) {
var max = (val > prev.max) ? val : prev.max
var diff = (max - val > prev.diff) ? max - val : prev.diff
return {max: max, diff: diff}
}, {max: 0, diff: 0})
return result.diff
}
Related
So I am trying to solve this task "Digit Difference Sort" on Codefights
Given an array of integers, sort its elements by the difference of their largest and smallest digits.
In the case of a tie, that with the larger index in the array should come first.
Example
For a = [152, 23, 7, 887, 243], the output should be digitDifferenceSort(a) = [7, 887, 23, 243, 152].
Here are the differences of all the numbers:
152: difference = 5 - 1 = 4;
23: difference = 3 - 2 = 1;
7: difference = 7 - 7 = 0;
887: difference = 8 - 7 = 1;
243: difference = 4 - 2 = 2.
23 and 887 have the same difference, but 887 goes after 23 in a, so in the sorted array it comes first.
I have an issue with two numbers having the same difference. Here's what I wrote so far:
int[] digitDifferenceSort(int[] a) {
return a.OrderBy(x => difference(x)).ToArray();
}
int difference(int x)
{
int min = 9, max = 0;
do
{
int tmp = x % 10;
min = Math.Min(min, tmp);
max = Math.Max(max, tmp);
} while ((x /= 10) > 0);
return max - min;
}
Didn't do much (for example the output is still [7, 23, 887, 243, 152] rather than [7, 887, 23, 243, 152])
How do I make element with larger index come first in result? What should I use instead of OrderBy?
I don't consider your difference method, i assume it works fine.
To your question: you have to keep revered order of the array (that the items with the same difference arrive will be sorted reverse). To do it, you could just reverse you input array: all items with not identical difference will be ordered correctly, and with the same differece will be ordered reversed:
int[] digitDifferenceSort(int[] a)
{
return a.Reverse().OrderBy(x => difference(x)).ToArray();
}
Following is my code for the above question digit difference sort. I am also getting output when running in Eclipse but when I paste the code on code signal it gives me a null pointer exception.
package NormalPrograms;
import java.util.ArrayList;
import java.util.Collections;
public class DigitDifferenceSort {
// For index wise sorting in descending order
public static int[] sortingnumberindexwise(int[] a, ArrayList<Integer> index) {
int k = 0;
int[] res = new int[index.size()];
int[] finalres = new int[index.size()];
for (int i = a.length - 1; i >= 0; i--) {
for (int j = 0; j < index.size(); j++) {
if (a[i] == (int) index.get(j)) {
res[k] = i;
index.remove(j);
k++;
break;
}
}
}
int g = 0;
k = 0;
for (int i = 0; i < res.length; i++) {
finalres[g] = a[res[k]];
g++;
k++;
}
return finalres;
}
public static int[] finddigitDifferenceandSort(int[] p) {
int[] finres = new int[p.length];
for (int i = 0; i < finres.length; i++) {
finres[i] = p[i];
}
// This finres array act as an temp array and reused to make final result array
int digit = 0;
ArrayList<Integer> A = new ArrayList<Integer>();
ArrayList<ArrayList<Integer>> B = new ArrayList<ArrayList<Integer>>();
for (int i = 0; i < 10; i++) {
B.add(new ArrayList<Integer>());
}
for (int i = 0; i < p.length; i++) {
int temp = 0;
temp = p[i];
while (p[i] > 0) {
digit = p[i] % 10;
p[i] /= 10;
A.add(digit);
}
int b = Collections.max(A);
int c = Collections.min(A);
int diff = b - c;
B.get(diff).add(temp);
A.clear();
}
for (int i = 0; i < B.size(); i++) {
if (B.get(i).size() > 1) {
ArrayList<Integer> C = new ArrayList<Integer>();
for (int k = 0; k < B.get(i).size(); k++) {
C.add(B.get(i).get(k));
}
B.get(i).clear();
for (int j : sortingnumberindexwise(finres, C)) {
B.get(i).add(j);
}
} else {
continue;
}
}
int k = 0;
for (int i = 0; i < B.size(); i++) {
for (int j = 0; j < B.get(i).size(); j++) {
if (B.get(i).size() == 0)
continue;
else {
finres[k] = B.get(i).get(j);
k++;
}
}
}
return finres;
}
public static void main(String[] args) {
int[] a = { 12, 21, 1, 1, 1, 2, 2, 3 };
for (int i : finddigitDifferenceandSort(a)) {
System.out.print(i + " ");
}
}
}
I am solving this problem, in which they ask for the index of the first Fibonacci number of 1000 digits, and my first idea was something similar to:
BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;
int currentIndex = 2;
while (x.NoOfDigits < 1000)
{
tmp = x + y;
y = x;
x = tmp;
currentIndex++;
}
return currentIndex;
However, as far as I can tell, there is no method for counting the number of digits of a BigInteger. Is this true? One way of circumventing it is to use the .ToString().Length method of a BigInteger, but I'm told that string processing is slow.
A BigInteger also has a .ToByteArray(), and I thought of converting a BigInteger to a byte array and checking the length of that array - but I don't think that this uniquely determines the number of digits in the BigInteger.
For what it's worth, I implemented another way of solving it, which is manually storing the Fibonacci numbers in array, and which stops as soon as the array is full, and I compared this to the .ToString-based method, which is about 2.5 times slower, but the first method takes 0.1 second, which also seems like a long time.
Edit: I've tested the two suggestions in the answers below (the one with BigInteger.Log and the one with MaxLimitMethod). I get the following run times:
Original method: 00:00:00.0961957
StringMethod: 00:00:00.1535350
BigIntegerLogMethod: 00:00:00.0387479
MaxLimitMethod: 00:00:00.0019509
Program
using System;
using System.Collections.Generic;
using System.Numerics;
using System.Diagnostics;
class Program
{
static void Main(string[] args)
{
Stopwatch clock = new Stopwatch();
clock.Start();
int index1 = Algorithms.IndexOfNDigits(1000);
clock.Stop();
var elapsedTime1 = clock.Elapsed;
Console.WriteLine(index1);
Console.WriteLine("Original method: {0}",elapsedTime1);
Console.ReadKey();
clock.Reset();
clock.Start();
int index2 = Algorithms.StringMethod(1000);
clock.Stop();
var elapsedTime2 = clock.Elapsed;
Console.WriteLine(index2);
Console.WriteLine("StringMethod: {0}", elapsedTime2);
Console.ReadKey();
clock.Reset();
clock.Start();
int index3 = Algorithms.BigIntegerLogMethod(1000);
clock.Stop();
var elapsedTime3 = clock.Elapsed;
Console.WriteLine(index3);
Console.WriteLine("BigIntegerLogMethod: {0}", elapsedTime3);
Console.ReadKey();
clock.Reset();
clock.Start();
int index4 = Algorithms.MaxLimitMethod(1000);
clock.Stop();
var elapsedTime4 = clock.Elapsed;
Console.WriteLine(index4);
Console.WriteLine("MaxLimitMethod: {0}", elapsedTime4);
Console.ReadKey();
}
}
static class Algorithms
{
//Find the index of the first Fibonacci number of n digits
public static int IndexOfNDigits(int n)
{
if (n == 1) return 1;
int[] firstNumber = new int[n];
int[] secondNumber = new int[n];
firstNumber[0] = 1;
secondNumber[0] = 1;
int currentIndex = 2;
while (firstNumber[n-1] == 0)
{
int carry = 0, singleSum = 0;
int[] tmp = new int[n]; //Placeholder for the sum
for (int i = 0; i<n; i++)
{
singleSum = firstNumber[i] + secondNumber[i];
if (singleSum >= 10) carry = 1;
else carry = 0;
tmp[i] += singleSum % 10;
if (tmp[i] >= 10)
{
tmp[i] = 0;
carry = 1;
}
int countCarries = 0;
while (carry == 1)
{
countCarries++;
if (tmp[i + countCarries] == 9)
{
tmp[i + countCarries] = 0;
tmp[i + countCarries + 1] += 1;
carry = 1;
}
else
{
tmp[i + countCarries] += 1;
carry = 0;
}
}
}
for (int i = 0; i < n; i++ )
{
secondNumber[i] = firstNumber[i];
firstNumber[i] = tmp[i];
}
currentIndex++;
}
return currentIndex;
}
public static int StringMethod(int n)
{
BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;
int currentIndex = 2;
while (x.ToString().Length < n)
{
tmp = x + y;
y = x;
x = tmp;
currentIndex++;
}
return currentIndex;
}
public static int BigIntegerLogMethod(int n)
{
BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;
int currentIndex = 2;
while (Math.Floor(BigInteger.Log10(x) + 1) < n)
{
tmp = x + y;
y = x;
x = tmp;
currentIndex++;
}
return currentIndex;
}
public static int MaxLimitMethod(int n)
{
BigInteger maxLimit = BigInteger.Pow(10, n - 1);
BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;
int currentIndex = 2;
while (x.CompareTo(maxLimit) < 0)
{
tmp = x + y;
y = x;
x = tmp;
currentIndex++;
}
return currentIndex;
}
}
Provided that x > 0
int digits = (int)Math.Floor(BigInteger.Log10(x) + 1);
will get the number of digits.
Out of curiosity, I tested the
int digits = x.ToString().Length;
approach. For 100 000 000 iterations, it's 3 times slower than the Log10 solution.
Expanding on my comment--instead of testing based on number of digits, test based on exceeding a constant that has the upper limit of the problem:
public static int MaxLimitMethod(int n)
{
BigInteger maxLimit = BigInteger.Pow(10, n);
BigInteger x = 1;
BigInteger y = 1;
BigInteger tmp = 0;
int currentIndex = 2;
while (x.CompareTo(maxLimit) < 0)
{
tmp = x + y;
y = x;
x = tmp;
currentIndex++;
}
return currentIndex;
}
This should result in a significant performance increase.
UPDATE:
This is an even quicker method on .NET 5 (since GetBitLength() is required):
private static readonly double exponentConvert = Math.Log10(2);
private static readonly BigInteger _ten = 10;
public static int CountDigits(BigInteger value)
{
if (value.IsZero)
return 1;
value = BigInteger.Abs(value);
if (value.IsOne)
return 1;
long numBits = value.GetBitLength();
int base10Digits = (int)(numBits * exponentConvert).Dump();
var reference = BigInteger.Pow(_ten, base10Digits);
if (value >= reference)
base10Digits++;
return base10Digits;
}
The slowest part of this algorithm for large values is the BigInteger.Pow() operation. I have optimized the CountDigits() method in Singulink.Numerics.BigIntegerExtensions with a cache that holds powers of 10, so check that out if you are interested in the fastest possible implementation. It caches powers up to exponents of 1023 by default but if you want to trade memory usage for faster performance on even larger values you can increase the max cached exponent by calling BigIntegerPowCache.GetCache(10, maxSize) where maxSize = maxExponent + 1.
On an i7-3770 CPU, this library takes 350ms to get the digit count for 10 million BigInteger values (single-threaded) when the digit count <= the max cached exponent.
ORIGINAL ANSWER:
The accepted answer is unreliable, as indicated in the comments. This method works for all numbers:
private static int CountDigits(BigInteger value)
{
if (value.IsZero)
return 1;
value = BigInteger.Abs(value);
if (value.IsOne)
return 1;
int exp = (int)Math.Ceiling(BigInteger.Log10(value));
var test = BigInteger.Pow(10, exp);
return value >= test ? exp + 1 : exp;
}
I have five long integers p, q, s, m and x. An array numbers[] is created by the following formula.
numbers[0] = s;
for(int i=1; i<numbers.Length;i++){
numbers[i] = (p * numbers[i-1] + q) % m;
}
The first value of numbers (numbers[0]) is s.
What is the most efficient way to find index j where i < j and |numbers[j] - numbers[i]| <= x or |numbers[j] - numbers[i]| >= m-x.
For instance, in a case where p = 3, q= 7, s= 1, m= 29 en x= 1 the array will be:
numbers[0] = 1, numbers[1] = 10, numbers[2] = 8 and numbers[3] = 2.
In this case index j would be 3, because numbers[3] - numbers[0]<=x, because x is 1.
I thought about using something such as a variant of counting sort or radix sort but I can't get anything to work.
As i < j, then you need to grant that numbers has a length of at least 2.
You could do two nested loops, the outer one ranging from j = 1 to numbers.Length - 1 (granting the possible solution to be the smallest j) to i = 0 to i < j.
Then you compare both positions according your specs. If true, return j. If it finishes both loops, then there is no solution.
Edit: Code Sample
public int GetSmallestIndex(long[] numbers, long x, long m)
{
if (numbers.Length >= 2)
{
for (int j = 1; j < numbers.Length; j++)
{
for (int i = 0; i < j; i++)
{
long diff = Math.Abs(numbers[j] - numbers[i]);
if (diff <= x || diff >= m - x)
return j;
}
}
}
return -1; //If no solution is found, return -1 as convention
}
The only way to find out if something is more efficient is to benchmark it by using the StopWatch in System.Diagnostics.
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
namespace ConsoleApplication1
{
class Program
{
public static void Main()
{
var stopWatch = new Stopwatch();
stopWatch.Start();
const long x = 1;
var numbers = new long[] {3, 7, 1, 29};
var theSmallestIndex = SmallestIndex(x, numbers);
stopWatch.Stop();
Console.WriteLine("Elapsed Time: {0}", stopWatch.Elapsed);
Console.WriteLine("Smallest Index: {0}", theSmallestIndex);
Console.ReadKey();
}
public static long SmallestIndex(long x, long[] numbers)
{
var values = ValuesMinusTheValueOfPreviousIndex(x, numbers.ToList());
var smallest = values.Values.OrderBy(n => n).FirstOrDefault();
var result = values.Where(n => n.Value.Equals(smallest));
return result.FirstOrDefault().Key;
}
public static Dictionary<int, long> ValuesMinusTheValueOfPreviousIndex(long x, List<long> numbers)
{
var results = new Dictionary<int, long>();
foreach (var number in numbers)
{
var index = numbers.IndexOf(number);
var previousNumber = index > 0 ? numbers.ElementAt(index - 1) : 0;
var result = number - previousNumber;
results.Add(index, result);
}
return results;
}
}
}
EDIT: Added Math.Abs as you requested in the comments
long p = 3, q = 7, s = 1, m = 29, x = 1;
long[] numbers = new long[10];
numbers[0] = s;
for (int i = 1; i < numbers.Length; i++)
{
numbers[i] = (p * numbers[i - 1] + q) % m;
}
// Find the smallest index j in numbers, where i < j &&
// (numbers[j] - numbers[i] <= x || numbers[j] - numbers[i] >= m-x)
int smallestIndex = 0;
long comparison;
for (int j = 1; j < numbers.Length; j++)
{
for (int i = 0; i < j; i++)
{
comparison = Math.Abs(numbers[j] - numbers[i]);
if (comparison <= x || comparison >= m - x)
{
smallestIndex = j;
break;
}
}
if (smallestIndex != 0) break;
}
if (smallestIndex == 0)
{
// No result matches the conditions
}
else
{
// j is the smallest index matching the conditions
// Before using Abs, in the example j is 2, because 8 - 10 = -2, lesser than x
// Now using absolute values, In the example j is 3
}
Just got done with the latest Codility, passed it, but didnt get 100% on it
Here is the spec
A prefix of a string S is any leading contiguous part of S. For example, "c" and "cod" are prefixes of the string "codility". For simplicity, we require prefixes to be non-empty.
The product of prefix P of string S is the number of occurrences of P multiplied by the length of P. More precisely, if prefix P consists of K characters and P occurs exactly T times in S, then the product equals K * T.
For example, S = "abababa" has the following prefixes:
"a", whose product equals 1 * 4 = 4,
"ab", whose product equals 2 * 3 = 6,
"aba", whose product equals 3 * 3 = 9,
"abab", whose product equals 4 * 2 = 8,
"ababa", whose product equals 5 * 2 = 10,
"ababab", whose product equals 6 * 1 = 6,
"abababa", whose product equals 7 * 1 = 7.
The longest prefix is identical to the original string. The goal is to choose such a prefix as maximizes the value of the product. In above example the maximal product is 10.
In this problem we consider only strings that consist of lower-case English letters (a−z).
So basically, it is a string traverse problem. I was able to pass all the validation parts, but I lost on the time. Here is what I wrote
int Solution(string S)
{
int finalCount = 0;
for (int i = 0; i <= S.Length - 1; i++)
{
string prefix = S.Substring(0, i + 1);
int count = 0;
for (int j = 0; j <= S.Length - 1; j++)
{
if (prefix.Length + j <= S.Length)
{
string newStr = S.Substring(j, prefix.Length);
if (newStr == prefix)
{
count++;
}
}
if (j == S.Length - 1)
{
int product = count * prefix.Length;
if (product > finalCount)
{
finalCount = product;
}
}
}
}
return finalCount;
}
I know that the nested loop is killing me, but I cannot think of a way to traverse the "sections" of the string without adding the other loop.
Any help would be appreciated.
Naive brute-force solution takes O(N ** 3) time. Choose length from 1 to N, get a prefix of its length and count occurences by brute-force searching. Сhoosing length takes O(N) time and brute force takes O(N ** 2) time, totally O(N ** 3).
If you use KMP or Z-algo, you can find occurences in O(N) time, so the whole solution will be O(N ** 2) time.
And you can precalc occurences, so it will take O(N) + O(N) = O(N) time solution.
vector<int> z_function(string &S); //z-function, z[0] = S.length()
vector<int> z = z_function(S);
//cnt[i] - count of i-length prefix occurrences of S
for (int i = 0; i < n; ++i)
++cnt[z[i]];
//if cnt[i] is in S, cnt[i - 1] will be in S
int previous = 0;
for (int i = n; i > 0; --i) {
cnt[i] += previous;
previous = cnt[i];
}
Here's the blog post, explaining all O(N ** 3), O(N ** 2), O(N) solutions.
my effort was as follows trying to eliminate unnecessary string compares, i read isaacs blog but it is in c how would this translate to c#, i even went as far as using arrays everywhere to avoid the string immutability factor but there was no improvement
int Rtn = 0;
int len = S.Length;
if (len > 1000000000)
return 1000000000;
for (int i = 1; i <= len; i++)
{
string tofind = S.Substring(0, i);
int tofindlen = tofind.Length;
int occurencies = 0;
for (int ii = 0; ii < len; ii++)
{
int found = FastIndexOf(S, tofind.ToCharArray(), ii);
if (found != -1)
{
if ((found == 0 && tofindlen != 1) || (found >= 0))
{
ii = found;
}
++occurencies ;
}
}
if (occurencies > 0)
{
int total = occurencies * tofindlen;
if (total > Rtn)
{
Rtn = total;
}
}
}
return Rtn;
}
static int FastIndexOf(string source, char[] pattern, int start)
{
if (pattern.Length == 0) return 0;
if (pattern.Length == 1) return source.IndexOf(pattern[0], start);
bool found;
int limit = source.Length - pattern.Length + 1 - start;
if (limit < 1) return -1;
char c0 = pattern[0];
char c1 = pattern[1];
// Find the first occurrence of the first character
int first = source.IndexOf(c0, start, limit);
while ((first != -1) && (first + pattern.Length <= source.Length))
{
if (source[first + 1] != c1)
{
first = source.IndexOf(c0, ++first);
continue;
}
found = true;
for (int j = 2; j < pattern.Length; j++)
if (source[first + j] != pattern[j])
{
found = false;
break;
}
if (found) return first;
first = source.IndexOf(c0, ++first);
}
return -1;
}
I only got a 43... I like my code though! Same script in javascript got 56 for what it's worth.
using System;
class Solution
{
public int solution(string S)
{
int highestCount = 0;
for (var i = S.Length; i > 0; i--)
{
int occurs = 0;
string prefix = S.Substring(0, i);
int tempIndex = S.IndexOf(prefix) + 1;
string tempString = S;
while (tempIndex > 0)
{
tempString = tempString.Substring(tempIndex);
tempIndex = tempString.IndexOf(prefix);
tempIndex++;
occurs++;
}
int product = occurs * prefix.Length;
if ((product) > highestCount)
{
if (highestCount > 1000000000)
return 1000000000;
highestCount = product;
}
}
return highestCount;
}
}
This works better
private int MaxiumValueOfProduct(string input)
{
var positions = new List<int>();
int max = 0;
for (int i = 1; i <= input.Length; i++)
{
var subString = input.Substring(0, i);
int position = 0;
while ((position < input.Length) &&
(position = input.IndexOf(subString, position, StringComparison.OrdinalIgnoreCase)) != -1)
{
positions.Add(position);
++position;
}
var currentMax = subString.Length * positions.Count;
if (currentMax > max) max = currentMax;
positions.Clear();
}
return max;
}
I'm looking to assess similarity (including case) between two strings and give a value between 0 and 1.
I tried the Levenshtein distance implementation but it only gives integers and does not compare inner alphabets.
For e.g. comparing "ABCD" and "Abcd" gives distance of 3 and "AOOO" also gives a distance of 3 but clearly "Abcd" is better match than "AOOO".
So compared to "ABCD" I want "ABcd" to be most similar then "Abcd" then "AOOO" then "AOOOO"
I've also looked here but I am not looking for a variable length algorithm.
Thanks
Try something like this
double d = (LevenshteinDist(s, t) + LevenshteinDist(s.ToLower(), t.ToLower())) /
2.0 * Math.Max(s.Length, t.Length);
If you want to give less importance to case differences than letter differences, you can give different weights to the terms
double d = (0.15*LevenshteinDist(s, t) +
0.35*LevenshteinDist(s.ToLower(), t.ToLower())) /
Math.Max(s.Length, t.Length);
Note that the weights sum up to 0.5, thus makting the division by 2.0 obsolete.
Adapt Levenshtein Distance with a custom table T. Let the cost of insertion = 1. The cost of deletion also 1. Let T(c,d) denote the penalty of replacing c with d. T(c,c) should be = 0. T(c,d) should be <= 2.
Define Max(n,m) be the maximum theoretical distance of strings of length n and m. Obviously, Max(n,m) = n+m.
Define Distance(s,t) be the cost of changing s to t divided by Max(s,t). There you go.
Be careful in defining T so that the definition obeys distance axioms:
Distance(s,s) = 0
Distance(s,t) = Distance(t,s)
Distance(s,t) <= Distance(s,u) + Distance(u,t)
Then it will be more useful in more situations.
bool check(string[] a, string s)
{
for (int i = 0; i < a.Length; i++)
if (s == a[i])
return true;
return false;
}
public double simi(string string1, string string2)
{
int sub1 = 0;
int sub2 = 0;
string[] sp1 = new string[string1.Length - 1];
string[] sp2 = new string[string2.Length - 1];
string[] sp3 = new string[string1.Length - 1];
string[] sp4 = new string[string2.Length - 1];
for (int i = 0; i < string1.Length - 1; i++)
{
string x = "";
x = string1.Substring(i, 2);
sp1[sub1] = x;
++sub1;
}
for (int i = 0; i < string2.Length - 1; i++)
{
string x = "";
x = string2.Substring(i, 2);
sp2[sub2] = x;
++sub2;
}
int j = 0, k = 0;
for (int i = 0; i < sp1.Length; i++)
if (check(sp3, sp1[i]) == true)
{
continue;
}
else
{
sp3[j] = sp1[i];
j++;
}
for (int i = 0; i < sp2.Length; i++)
if (check(sp4, sp2[i]) == true)
{
continue;
}
else
{
sp4[k] = sp2[i];
k++;
}
Array.Resize(ref sp3, j);
Array.Resize(ref sp4, k);
Array.Sort<string>(sp3);
Array.Sort<string>(sp4);
int n = 0;
for (int i = 0; i < sp3.Length; i++)
{
if (check(sp4, sp3[i]))
{
n++;
}
}
double resulte;
int l1 = sp3.Length;
int l2 = sp4.Length;
resulte = ((2.0 * Convert.ToDouble(n)) / Convert.ToDouble(l1 + l2)) * 100;
return resulte;
}