Count of substrings of a binary string containing K ones
Last Updated :
20 Jul, 2022
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:
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).
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int countOfSubstringWithKOnes(string s, int K)
{
int N = s.length();
int res = 0;
int countOfOne = 0;
int freq[N + 1] = {0};
freq[0] = 1;
for (int i = 0; i < N; i++) {
countOfOne += (s[i] - '0');
if (countOfOne >= K) {
res += freq[countOfOne - K];
}
freq[countOfOne]++;
}
return res;
}
int main()
{
string s = "10010";
int K = 1;
cout << countOfSubstringWithKOnes(s, K) << endl;
return 0;
}
|
Java
import java.io.*;
public class GFG {
static int countOfSubstringWithKOnes(
String s, int K)
{
int N = s.length();
int res = 0;
int countOfOne = 0;
int []freq = new int[N+1];
freq[0] = 1;
for (int i = 0; i < N; i++) {
countOfOne += (s.charAt(i) - '0');
if (countOfOne >= K) {
res += freq[countOfOne - K];
}
freq[countOfOne]++;
}
return res;
}
static public void main (String[] args)
{
String s = "10010";
int K = 1;
System.out.println(
countOfSubstringWithKOnes(s, K));
}
}
|
Python3
def countOfSubstringWithKOnes(s, K):
N = len(s)
res = 0
countOfOne = 0
freq = [0 for i in range(N + 1)]
freq[0] = 1
for i in range(0, N, 1):
countOfOne += ord(s[i]) - ord('0')
if (countOfOne >= K):
res += freq[countOfOne - K]
freq[countOfOne] += 1
return res
if __name__ == '__main__':
s = "10010"
K = 1
print(countOfSubstringWithKOnes(s, K))
|
C#
using System;
public class GFG {
static int countOfSubstringWithKOnes(
string s, int K)
{
int N = s.Length;
int res = 0;
int countOfOne = 0;
int []freq = new int[N+1];
freq[0] = 1;
for (int i = 0; i < N; i++) {
countOfOne += (s[i] - '0');
if (countOfOne >= K) {
res += freq[countOfOne - K];
}
freq[countOfOne]++;
}
return res;
}
static public void Main ()
{
string s = "10010";
int K = 1;
Console.WriteLine(
countOfSubstringWithKOnes(s, K));
}
}
|
PHP
<?php
function countOfSubstringWithKOnes($s, $K)
{
$N = strlen($s);
$res = 0;
$countOfOne = 0;
$freq = array();
for ($i = 0; $i <= $N; $i++)
$freq[$i] = 0;
$freq[0] = 1;
for ($i = 0; $i < $N; $i++)
{
$countOfOne += ($s[$i] - '0');
if ($countOfOne >= $K)
{
$res = $res + $freq[$countOfOne - $K];
}
$freq[$countOfOne]++;
}
return $res;
}
$s = "10010";
$K = 1;
echo countOfSubstringWithKOnes($s, $K) ,"\n";
?>
|
Javascript
<script>
function countOfSubstringWithKOnes(s, K)
{
let N = s.length;
let res = 0;
let countOfOne = 0;
let freq = new Array(N + 1);
freq.fill(0);
freq[0] = 1;
for(let i = 0; i < N; i++)
{
countOfOne += (s[i] - '0');
if (countOfOne >= K)
{
res += freq[countOfOne - K];
}
freq[countOfOne]++;
}
return res;
}
let s = "10010";
let K = 1;
document.write(countOfSubstringWithKOnes(s, K));
</script>
|
Time Complexity: O(N).
Auxiliary Space: O(N).
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