Maximize sum by choosing Array aspect to left of every ‘1’ of a Binary String

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Given a binary string S and an array arr[] every of dimension N, we will choose any aspect from the Array which is to the left of (or on the similar place) the indices of ‘1’s within the given binary string. The duty is to seek out the utmost potential sum.

Examples:

Enter: arr[] = {20, 10, 30, 9, 20, 9}, string S = “011011”, N = 6
Output: 80
Clarification: Decide 20, 10, 30 and 20 in Sum, so, Sum = 80.

Enter:  arr[] = {30, 20, 10}, string S = “000”, N = 3.
Output: 0

Strategy: The given drawback might be solved through the use of a precedence queue based mostly on the next thought:

Say there are Okay occurrences of ‘1’ in string S. It may be seen that we will organize the characters in a method such that we will choose the Okay most parts from the array that are to the left of the final prevalence of ‘1’ in S. So we will use a precedence queue to get these Okay most parts.

Observe the steps to unravel this drawback:

  • Initialize variable Sum = 0, Cnt = 0.
  • Create a precedence queue (max heap) and traverse from i = 0 to N-1:
    • If S[i] is ‘1’, increment Cnt by 1.
    • Else, whereas Cnt > 0, add the topmost aspect of the precedence queue and decrement Cnt by 1.
    • Push the ith aspect of the array into the precedence queue.
  • After executing the loop, whereas Cnt > 0, add the topmost aspect of the precedence queue and decrement Cnt by 1.
  • Eventually, return the Sum because the required reply.

Beneath is the implementation of the above method.

C++

  

#embrace <bits/stdc++.h>

utilizing namespace std;

  

int findMaxSum(int* arr, string s, int n)

{

    

    int Cnt = 0, Sum = 0;

  

    priority_queue<int> pq;

  

    

    for (int i = 0; i < n; i++) {

        if (s[i] == '1') {

            Cnt++;

        }

        else {

            whereas (Cnt != 0) {

                Sum += pq.high();

                pq.pop();

                Cnt--;

            }

        }

  

        

        pq.push(arr[i]);

    }

  

    whereas (Cnt != 0) {

        Sum += pq.high();

        pq.pop();

        Cnt--;

    }

  

    

    return Sum;

}

  

int fundamental()

{

    int N = 6;

    string S = "011011";

    int arr[] = { 20, 10, 30, 9, 20, 9 };

  

    

    cout << findMaxSum(arr, S, N) << endl;

  

    return 0;

}

Time Complexity: O(N * log N)
Auxiliary Area: O(N)

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