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Count Substrings that can be made of length 1 by replacing “01” or “10” with 1 or 0

Kashish Kumar
Published: August 2, 2022

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Given a binary string S of length N, the task is to find the number of pairs of integers [L, R] 1 ≤ L < R ≤ N such that S[L . . . R] (the substring of S from L to R) can be reduced to 1 length string by replacing substrings “01” or “10” with “1” and “0” respectively.

Examples:

Input: S = “0110”
Output: 4
Explanation: The 4 substrings are 01, 10, 110, 0110.

Input: S = “00000”
Output: 0

 

Approach: The solution is based on the following mathematical idea:

We can solve this based on the exclusion principle. Instead of finding possible pairs find the number of impossible cases and subtract that from all possible substrings (i.e. N*(N+1)/2 ).

How to find impossible cases?

When s[i] and s[i-1] are same, then after reduction it will either become “00” or “11”. In both cases, the substring cannot be reduced to length 1. So substring from 0 to i, from 1 to i, . . . cannot be made to have length 1. That count of substrings is i.

Follow the below steps to solve the problem:

  • Initialize answer ans = N * (N + 1) / 2
  • Run a loop from i = 1 to N – 1
    • If S[i] is equal to S[i – 1], then subtract i from ans.
  • Return ans – N (because there are N substrings having length 1).

Below is the implementation of the above approach.

C++

  

#include <bits/stdc++.h>

#define ll long long

using namespace std;

  

ll find(string Str)

{

  

    ll n = Str.size();

  

    ll ans = n * (n + 1) / 2;

  

    for (ll i = 1; i < n; i++) {

        if (Str[i] == Str[i - 1])

            ans -= i;

    }

  

    return ans - n;

}

  

int main()

{

    string S = "0110";

  

    

    cout << find(S) << endl;

    return 0;

}

Time Complexity: O(N)
Auxiliary Space: O(1)

Source: www.geeksforgeeks.org