# Count of substrings of a binary string containing K ones

Given a binary string of length N and an integer K, we need to find out how many substrings of this string are exist which contains exactly K ones.

**Examples:**

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Input : s = “10010” K = 1 Output : 9 The 9 substrings containing one 1 are, “1”, “10”, “100”, “001”, “01”, “1”, “10”, “0010” and “010”

In this problem we need to find count of substrings which contains exactly K ones or in other words sum of digits in those substring is K. We first create a prefix sum array and loop over that and stop when sum value is greater than or equal to K. Now if sum at current index is (K + a) then we know that substring sum, from all those indices where sum is (a), till current index will be K, so count of indices having sum (a), will be added to result. This procedure is explained with an example below,

string s = “100101” K = 2 prefix sum array = [1, 1, 1, 2, 2, 3] So, at index 3, we have prefix sum 2, Now total indices from where sum is 2, is 1 so result = 1 Substring considered = [“1001”] At index 4, we have prefix sum 2, Now total indices from where sum is 2, is 1 so result = 2 Substring considered = [“1001”, “10010”] At index 5, we have prefix sum 3, Now total indices from where sum is 2, is 3 so result = 5 Substring considered = [“1001”, “10010”, “00101”, “0101”, “101”]

So we need to track two things, prefix sum and frequency of particular sum. In below code, instead of storing complete prefix sum, only prefix sum at current index is stored using one variable and frequency of sums in stored in an array. Total time complexity of solution is O(N).

## C++

`// C++ program to find count of substring containing` `// exactly K ones` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// method returns total number of substring having K ones` `int` `countOfSubstringWithKOnes(string s, ` `int` `K)` `{` ` ` `int` `N = s.length();` ` ` `int` `res = 0;` ` ` `int` `countOfOne = 0;` ` ` `int` `freq[N + 1] = {0};` ` ` `// initialize index having zero sum as 1` ` ` `freq[0] = 1;` ` ` `// loop over binary characters of string` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// update countOfOne variable with value` ` ` `// of ith character` ` ` `countOfOne += (s[i] - ` `'0'` `);` ` ` `// if value reaches more than K, then` ` ` `// update result` ` ` `if` `(countOfOne >= K) {` ` ` `// add frequency of indices, having` ` ` `// sum (current sum - K), to the result` ` ` `res += freq[countOfOne - K];` ` ` `}` ` ` `// update frequency of one's count` ` ` `freq[countOfOne]++;` ` ` `}` ` ` `return` `res;` `}` `// Driver code to test above methods` `int` `main()` `{` ` ` `string s = ` `"10010"` `;` ` ` `int` `K = 1;` ` ` `cout << countOfSubstringWithKOnes(s, K) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to find count of substring` `// containing exactly K ones` `import` `java.io.*;` `public` `class` `GFG {` ` ` `// method returns total number of` ` ` `// substring having K ones` ` ` `static` `int` `countOfSubstringWithKOnes(` ` ` `String s, ` `int` `K)` ` ` `{` ` ` `int` `N = s.length();` ` ` `int` `res = ` `0` `;` ` ` `int` `countOfOne = ` `0` `;` ` ` `int` `[]freq = ` `new` `int` `[N+` `1` `];` ` ` ` ` `// initialize index having zero` ` ` `// sum as 1` ` ` `freq[` `0` `] = ` `1` `;` ` ` ` ` `// loop over binary characters` ` ` `// of string` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` ` ` `// update countOfOne variable` ` ` `// with value of ith character` ` ` `countOfOne += (s.charAt(i) - ` `'0'` `);` ` ` ` ` `// if value reaches more than` ` ` `// K, then update result` ` ` `if` `(countOfOne >= K) {` ` ` ` ` `// add frequency of indices,` ` ` `// having sum (current sum - K),` ` ` `// to the result` ` ` `res += freq[countOfOne - K];` ` ` `}` ` ` ` ` `// update frequency of one's count` ` ` `freq[countOfOne]++;` ` ` `}` ` ` ` ` `return` `res;` ` ` `}` ` ` ` ` `// Driver code to test above methods` ` ` `static` `public` `void` `main (String[] args)` ` ` `{` ` ` `String s = ` `"10010"` `;` ` ` `int` `K = ` `1` `;` ` ` ` ` `System.out.println(` ` ` `countOfSubstringWithKOnes(s, K));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## Python3

`# Python 3 program to find count of` `# substring containing exactly K ones` `# method returns total number of` `# substring having K ones` `def` `countOfSubstringWithKOnes(s, K):` ` ` `N ` `=` `len` `(s)` ` ` `res ` `=` `0` ` ` `countOfOne ` `=` `0` ` ` `freq ` `=` `[` `0` `for` `i ` `in` `range` `(N ` `+` `1` `)]` ` ` `# initialize index having` ` ` `# zero sum as 1` ` ` `freq[` `0` `] ` `=` `1` ` ` `# loop over binary characters of string` ` ` `for` `i ` `in` `range` `(` `0` `, N, ` `1` `):` ` ` ` ` `# update countOfOne variable with` ` ` `# value of ith character` ` ` `countOfOne ` `+` `=` `ord` `(s[i]) ` `-` `ord` `(` `'0'` `)` ` ` `# if value reaches more than K,` ` ` `# then update result` ` ` `if` `(countOfOne >` `=` `K):` ` ` ` ` `# add frequency of indices, having` ` ` `# sum (current sum - K), to the result` ` ` `res ` `+` `=` `freq[countOfOne ` `-` `K]` ` ` ` ` `# update frequency of one's count` ` ` `freq[countOfOne] ` `+` `=` `1` ` ` ` ` `return` `res` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `s ` `=` `"10010"` ` ` `K ` `=` `1` ` ` `print` `(countOfSubstringWithKOnes(s, K))` `# This code is contributed by` `# Surendra_Gangwar` |

## C#

`// C# program to find count of substring` `// containing exactly K ones` `using` `System;` `public` `class` `GFG {` ` ` `// method returns total number of` ` ` `// substring having K ones` ` ` `static` `int` `countOfSubstringWithKOnes(` ` ` `string` `s, ` `int` `K)` ` ` `{` ` ` `int` `N = s.Length;` ` ` `int` `res = 0;` ` ` `int` `countOfOne = 0;` ` ` `int` `[]freq = ` `new` `int` `[N+1];` ` ` ` ` `// initialize index having zero` ` ` `// sum as 1` ` ` `freq[0] = 1;` ` ` ` ` `// loop over binary characters` ` ` `// of string` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` ` ` `// update countOfOne variable` ` ` `// with value of ith character` ` ` `countOfOne += (s[i] - ` `'0'` `);` ` ` ` ` `// if value reaches more than` ` ` `// K, then update result` ` ` `if` `(countOfOne >= K) {` ` ` ` ` `// add frequency of indices,` ` ` `// having sum (current sum - K),` ` ` `// to the result` ` ` `res += freq[countOfOne - K];` ` ` `}` ` ` ` ` `// update frequency of one's count` ` ` `freq[countOfOne]++;` ` ` `}` ` ` ` ` `return` `res;` ` ` `}` ` ` ` ` `// Driver code to test above methods` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `string` `s = ` `"10010"` `;` ` ` `int` `K = 1;` ` ` ` ` `Console.WriteLine(` ` ` `countOfSubstringWithKOnes(s, K));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find count` `// of substring containing` `// exactly K ones` `// method returns total number` `// of substring having K ones` `function` `countOfSubstringWithKOnes(` `$s` `, ` `$K` `)` `{` ` ` `$N` `= ` `strlen` `(` `$s` `);` ` ` `$res` `= 0;` ` ` `$countOfOne` `= 0;` ` ` `$freq` `= ` `array` `();` ` ` `for` `(` `$i` `= 0; ` `$i` `<= ` `$N` `; ` `$i` `++)` ` ` `$freq` `[` `$i` `] = 0;` ` ` `// initialize index` ` ` `// having zero sum as 1` ` ` `$freq` `[0] = 1;` ` ` ` ` `// loop over binary` ` ` `// characters of string` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$N` `; ` `$i` `++)` ` ` `{` ` ` `// update countOfOne` ` ` `// variable with value` ` ` `// of ith character` ` ` `$countOfOne` `+= (` `$s` `[` `$i` `] - ` `'0'` `);` ` ` `// if value reaches more` ` ` `// than K, then update result` ` ` `if` `(` `$countOfOne` `>= ` `$K` `)` ` ` `{` ` ` `// add frequency of indices,` ` ` `// having sum (current sum - K),` ` ` `// to the result` ` ` `$res` `= ` `$res` `+ ` `$freq` `[` `$countOfOne` `- ` `$K` `];` ` ` `}` ` ` `// update frequency` ` ` `// of one's count` ` ` `$freq` `[` `$countOfOne` `]++;` ` ` `}` ` ` `return` `$res` `;` `}` `// Driver code` `$s` `= ` `"10010"` `;` `$K` `= 1;` `echo` `countOfSubstringWithKOnes(` `$s` `, ` `$K` `) ,` `"\n"` `;` `// This code is contributed by m_kit` `?>` |

## Javascript

`<script>` `// Javascript program to find count of` `// substring containing exactly K ones` ` ` `// Method returns total number of` `// substring having K ones` `function` `countOfSubstringWithKOnes(s, K)` `{` ` ` `let N = s.length;` ` ` `let res = 0;` ` ` `let countOfOne = 0;` ` ` `let freq = ` `new` `Array(N + 1);` ` ` `freq.fill(0);` ` ` ` ` `// Initialize index having zero` ` ` `// sum as 1` ` ` `freq[0] = 1;` ` ` ` ` `// Loop over binary characters` ` ` `// of string` ` ` `for` `(let i = 0; i < N; i++)` ` ` `{` ` ` ` ` `// Update countOfOne variable` ` ` `// with value of ith character` ` ` `countOfOne += (s[i] - ` `'0'` `);` ` ` ` ` `// If value reaches more than` ` ` `// K, then update result` ` ` `if` `(countOfOne >= K)` ` ` `{` ` ` ` ` `// Add frequency of indices,` ` ` `// having sum (current sum - K),` ` ` `// to the result` ` ` `res += freq[countOfOne - K];` ` ` `}` ` ` ` ` `// Update frequency of one's count` ` ` `freq[countOfOne]++;` ` ` `}` ` ` `return` `res;` `}` `// Driver code` `let s = ` `"10010"` `;` `let K = 1;` `document.write(countOfSubstringWithKOnes(s, K));` `// This code is contributed by suresh07 ` `</script>` |

**Output:**

9

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