Given a binary string find the longest substring which contains 1’s more than 0’s.
Input : 1010 Output : 3 Substring 101 has 1 occurring more number of times than 0. Input : 101100 Output : 5 Substring 10110 has 1 occurring more number of times than 0.
A simple solution is to one by one consider all the substrings and check if that substring has a count of 1 more than 0. If the count is more than comparing its length with maximum length substring found till now. The time complexity of this solution is O(n^2).
An efficient solution is to use hashing. The idea is to find the sum of string traversed until now. Add 1 to the result if the current character is ‘1’ else subtract 1. Now the problem reduces to finding the largest subarray having a sum greater than zero. To find the largest subarray having a sum greater than zero, we check the value of the sum. If sum is greater than zero, then the largest subarray with a sum greater than zero is arr[0..i]. If the sum is less than zero, then find the size of subarray arr[j+1..i], where j is index up to 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 a 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 the length of largest subarray and that is obtained from first index occurrence.
Below is the implementation of above approach:
Time Complexity: O(n)
Auxiliary Space: O(n)
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