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Minimize sum of Array formed using given relation between adjacent elements

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  • Last Updated : 26 Apr, 2022
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Given a binary string S of length N, consisting of 0’s and 1’s, the task is to find the minimum sum of the array of non-negative integers of length N+1 created by following the below conditions:

  1. If the ith number in the given binary string is 0, then the (i + 1)th number in the array must be less than the ith number.
  2. If the ith number in the given binary string is 1, then the (i + 1)th number in the array must be greater than the ith number.

Examples:

Input: N = 3, S = “100”
Output: 3
Explanation:  We can create the array [0, 2, 1, 0]. 
So total sum will be 0 + 2 + 1 + 0  = 3. 
Hence, resultant array follows the conditions, and ‘3’ is the minimum value we can achieve.

Input: N = 3, S = “101”
Output: 2
Explanation: We can create the array [0, 1, 0, 1]. 
So total sum will be 0 + 1 + 0 + 1 = 2. 
Hence, resultant array follows the conditions, and ‘2’ is the minimum value we can achieve.

 

Approach: This problem can be solved by using greedy approach based on the following observation:

  • Consider we have K consecutive 1, in this case, last value will be at least K, as array would look something like this [0, 1, 2, …, K – 1, K], this would give us a minimum sum.
  • Same thing if we have K consecutive 0, in this case, array will look something like this  [K, K – 1, …, 2, 1, 0], hence our first value will be at least K.
  • Thus the ith element of the answer array will be the maximum among consecutive 1’s to its left and consecutive 0’s to its right. 

If we take a value greater than the maximum value, we will increase our sum, and hence the sum will not be minimum. If we take any less value than the maximum value, then one of the values in the array will become less than 0, which is a violation of the condition.

Follow the below steps to solve this problem:

  • Construct two arrays of length N + 1 (say arr1[] and arr2[] )and fill all the values as ‘0’.
  • Traverse from i = 0 to N – 1.
    • If S[i] value is 1 set arr1[i+1] = arr1[i]+1.
  • Traverse from i = N – 1 to 0.
    • If S[i] value is 0 set arr2[i] = arr2[i+1]+1.
  • Traverse in both the arrays from i = 0 to N:
    • Add the maximum of arr1[i] and arr2[i] to the answer.
  • Return the answer.

Below is the implementation of the above approach :

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the sum
long long minimumSum(string& s, int n)
{
    vector<int> arr1(n + 1, 0), arr2(n + 1, 0);
 
    // Finding maximum consecutive 1
    // to the left, for each index
    for (int i = 0; i < n; ++i) {
        if (s[i] == '1') {
            arr1[i + 1] = arr1[i] + 1;
        }
    }
 
    // Finding maximum consecutive
    // 0 to the right, for each index.
    for (int i = n - 1; i >= 0; --i) {
        if (s[i] == '0') {
            arr2[i] = arr2[i + 1] + 1;
        }
    }
 
    long long ans = 0;
 
    // Loop to find the sum
    for (int i = 0; i < n + 1; ++i) {
        ans += max(arr1[i], arr2[i]);
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int N = 3;
    string S = "101";
 
    // fzunction call
    cout << minimumSum(S, N);
    return 0;
}


Java




// Java code to implement the approach
import java.util.Arrays;
 
class GFG
{
 
  // Function to calculate the sum
  public static long minimumSum(String s, int n)
  {
    int arr1[] = new int[n + 1];
    int arr2[] = new int[n + 1];
    Arrays.fill(arr1, 0);
    Arrays.fill(arr2, 0);
 
    // Finding maximum consecutive 1
    // to the left, for each index
    for (int i = 0; i < n; ++i) {
      if (s.charAt(i) == '1') {
        arr1[i + 1] = arr1[i] + 1;
      }
    }
 
    // Finding maximum consecutive
    // 0 to the right, for each index.
    for (int i = n - 1; i >= 0; --i) {
      if (s.charAt(i) == '0') {
        arr2[i] = arr2[i + 1] + 1;
      }
    }
 
    long ans = 0;
 
    // Loop to find the sum
    for (int i = 0; i < n + 1; ++i) {
      ans += Math.max(arr1[i], arr2[i]);
    }
 
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 3;
    String S = "101";
 
    // function call
    System.out.println(minimumSum(S, N));
  }
}
 
// This code is contributed by phasing17


Python3




# Python3 code for the above approach
 
# Function to calculate the sum
def minimumSum(s, n):
    arr1 = [0] * (n + 1)
    arr2 = [0] * (n + 1)
 
    # Finding maximum consecutive 1
    # to the left, for each index
    for i in range(n):
        if s[i] == "1":
            arr1[i + 1] = arr1[i] + 1
 
    # Finding maximum consecutive
    # 0 to the right, for each index
    for i in range(n - 1, -1, -1):
        if s[i] == "0":
            arr2[i] = arr2[i + 1] + 1
    ans = 0
 
    # Loop to find the sum
    for i in range(n + 1):
        ans += max(arr1[i], arr2[i])
    return ans
 
# Driver code
N = 3
S = "101"
print(minimumSum(S, N))
 
# This code is contributed by phasing17


C#




// C# program to implement
// the above approach
using System;
class GFG
{
 
  // Function to calculate the sum
  public static long minimumSum(string s, int n)
  {
    int[] arr1 = new int[n + 1];
    int[] arr2 = new int[n + 1];
 
    for (int i = 0; i < n+1; i++) {
      arr1[i] = 0;
    }
    for (int i = 0; i < n+1; i++) {
      arr2[i] = 0;
    }
 
    // Finding maximum consecutive 1
    // to the left, for each index
    for (int i = 0; i < n; ++i) {
      if (s[i] == '1') {
        arr1[i + 1] = arr1[i] + 1;
      }
    }
 
    // Finding maximum consecutive
    // 0 to the right, for each index.
    for (int i = n - 1; i >= 0; --i) {
      if (s[i] == '0') {
        arr2[i] = arr2[i + 1] + 1;
      }
    }
 
    long ans = 0;
 
    // Loop to find the sum
    for (int i = 0; i < n + 1; ++i) {
      ans += Math.Max(arr1[i], arr2[i]);
    }
 
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 3;
    string S = "101";
 
    // function call
    Console.Write(minimumSum(S, N));
  }
}
 
// This code is contributed by code_hunt.


Javascript




<script>
       // JavaScript code for the above approach
 
 
       // Function to calculate the sum
       function minimumSum(s, n) {
           let arr1 = new Array(n + 1).fill(0), arr2 = new Array(n + 1).fill(0);
 
           // Finding maximum consecutive 1
           // to the left, for each index
           for (let i = 0; i < n; ++i) {
               if (s[i] == '1') {
                   arr1[i + 1] = arr1[i] + 1;
               }
           }
 
           // Finding maximum consecutive
           // 0 to the right, for each index.
           for (let i = n - 1; i >= 0; --i) {
               if (s[i] == '0') {
                   arr2[i] = arr2[i + 1] + 1;
               }
           }
 
           let ans = 0;
 
           // Loop to find the sum
           for (let i = 0; i < n + 1; ++i) {
               ans += Math.max(arr1[i], arr2[i]);
           }
 
           return ans;
       }
 
       // Driver Code
 
       let N = 3;
       let S = "101";
 
       // fzunction call
       document.write(minimumSum(S, N));
 
 
   // This code is contributed by Potta Lokesh
   </script>


Output

2

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


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