Given a binary string find the longest substring which contains 1’s more than 0’s.

Examples:

Input : 1010 Output : 3 Substring 101 has 1 occuring more number of times than 0. Input : 101100 Output : 5 Substring 10110 has 1 occuring more number of times than 0.

A **simple** solution is to one by one consider all the substrings and check if that substring has count of 1 more than 0. If count is more than compare its length with maximum length substring found till now. Time complexity of this solution is O(n^2).

An **efficient** solution is to use hashing. The idea is to find sum of string traversed until now. Add 1 to the result if current character is ‘1’ else subtract 1. Now the problem reduces to finding largest subarray having sum greater than zero. To find largest subarray having sum greater than zero, we check the value of sum. If sum is greater than zero, then largest subarray with sum greater than zero is arr[0..i]. If sum is less than zero, then find size of subarray arr[j+1..i], where j is index upto which sum of subarray arr[0..j] is sum -1 and j < i and compare that size with largest subarray size found so far. To find index j, store values of sum for arr[0..j] in hash table for all 0 <= j <= i. There might be possibility that a given value of sum repeats. In that case store only first index for which that sum is obtained as it is required to get length of largest subarray and that is obtained from first index occurrence.

Below is the implementation of above approach:

`// CPP program to find largest substring ` `// having count of 1s more than count ` `// count of 0s. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find longest substring ` `// having count of 1s more than count ` `// of 0s. ` `int` `findLongestSub(string bin) ` `{ ` ` ` `int` `n = bin.length(), i; ` ` ` ` ` `// To store sum. ` ` ` `int` `sum = 0; ` ` ` ` ` `// To store first occurrence of each ` ` ` `// sum value. ` ` ` `unordered_map<` `int` `, ` `int` `> prevSum; ` ` ` ` ` `// To store maximum length. ` ` ` `int` `maxlen = 0; ` ` ` ` ` `// To store current substring length. ` ` ` `int` `currlen; ` ` ` ` ` `for` `(i = 0; i < n; i++) { ` ` ` ` ` `// Add 1 if current character is 1 ` ` ` `// else subtract 1. ` ` ` `if` `(bin[i] == ` `'1'` `) ` ` ` `sum++; ` ` ` `else` ` ` `sum--; ` ` ` ` ` `// If sum is positive, then maximum ` ` ` `// length substring is bin[0..i] ` ` ` `if` `(sum > 0) { ` ` ` `maxlen = i + 1; ` ` ` `} ` ` ` ` ` `// If sum is negative, then maximum ` ` ` `// length substring is bin[j+1..i], where ` ` ` `// sum of substring bin[0..j] is sum-1. ` ` ` `else` `if` `(sum <= 0) { ` ` ` `if` `(prevSum.find(sum - 1) != prevSum.end()) { ` ` ` `currlen = i - prevSum[sum - 1]; ` ` ` `maxlen = max(maxlen, currlen); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Make entry for this sum value in hash ` ` ` `// table if this value is not present. ` ` ` `if` `(prevSum.find(sum) == prevSum.end()) ` ` ` `prevSum[sum] = i; ` ` ` `} ` ` ` ` ` `return` `maxlen; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `string bin = ` `"1010"` `; ` ` ` `cout << findLongestSub(bin); ` ` ` `return` `0; ` `} ` |

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**Output:**

3

**Time Complexity: ** O(n)

**Auxiliary Space: ** O(n)

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