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Maximize pair decrements required to reduce all array elements except one to 0
  • Difficulty Level : Medium
  • Last Updated : 04 Mar, 2021
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Given an array arr[] consisting of N distinct elements, the task is to find the maximum number of pairs required to be decreased by 1 in each step, such that N – 1 array elements are reduced to 0 and the remaining array element is a non-negative integer.

Examples:

Input: arr[] = {1, 2, 3}
Output: 3
Explanation: 
Decrease arr[1] and arr[2] by 1 modifies arr[] = {1, 1, 2} 
Decrease arr[1] and arr[2] by 1 modifies arr[] = {1, 0, 1} 
Decrease arr[0] and arr[2] by 1 modifies arr[] = {0, 0, 0} 
Therefore, the maximum number of decrements required is 3.
 

Input: arr[] = {1, 2, 3, 4, 5}
Output: 7

Approach: The problem can be solved Greedily. Follow the steps below to solve the problem:



  • Initialize a variable, say cntOp, to store maximum count of steps required to make (N – 1) elements of the array equal to 0.
  • Create a priority queue, say PQ, to store the array elements.
  • Traverse the array and insert the array elements into PQ.
  • Now repeatedly extract the top 2 elements from the priority queue, decrease the value of both the elements by 1, again insert both the elements in priority queue and increment the cntOp by 1. This process continues while (N – 1) element of the PQ becomes equal to 0.
  • Finally, print the value of cntOp

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count maximum number of steps
// to make (N - 1) array elements to 0
int cntMaxOperationToMakeN_1_0(int arr[], int N)
{
 
    // Stores maximum count of steps to make
    // (N - 1) elements equal to 0
    int cntOp = 0;
 
    // Stores array elements
    priority_queue<int> PQ;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Insert arr[i] into PQ
        PQ.push(arr[i]);
    }
 
    // Extract top 2 elements from the array
    // while (N - 1) array elements become 0
    while (PQ.size() > 1) {
 
        // Stores top element
        // of PQ
        int X = PQ.top();
 
        // Pop the top element
        // of PQ.
        PQ.pop();
 
        // Stores top element
        // of PQ
        int Y = PQ.top();
 
        // Pop the top element
        // of PQ.
        PQ.pop();
 
        // Update X
        X--;
 
        // Update Y
        Y--;
 
        // If X is not equal to 0
        if (X != 0) {
 
            // Insert X into PQ
            PQ.push(X);
        }
 
        // if Y is not equal
        // to 0
        if (Y != 0) {
 
            // Insert Y
            // into PQ
            PQ.push(Y);
        }
 
        // Update cntOp
        cntOp += 1;
    }
 
    return cntOp;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << cntMaxOperationToMakeN_1_0(arr, N);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to count maximum number of steps
// to make (N - 1) array elements to 0
static int cntMaxOperationToMakeN_1_0(int[] arr, int N)
{
     
    // Stores maximum count of steps to make
    // (N - 1) elements equal to 0
    int cntOp = 0;
     
    // Stores array elements
    PriorityQueue<Integer> PQ = new PriorityQueue<Integer>((a, b) -> b - a);
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Insert arr[i] into PQ
        PQ.add(arr[i]);
    }
     
    // Extract top 2 elements from the array
    // while (N - 1) array elements become 0
    while (PQ.size() > 1)
    {
         
        // Stores top element
        // of PQ
        int X = PQ.peek();
         
        // Pop the top element
        // of PQ.
        PQ.remove();
         
        // Stores top element
        // of PQ
        int Y = PQ.peek();
         
        // Pop the top element
        // of PQ.
        PQ.remove();
         
        // Update X
        X--;
         
        // Update Y
        Y--;
         
        // If X is not equal to 0
        if (X != 0)
        {
             
            // Insert X into PQ
            PQ.add(X);
        }
         
        // if Y is not equal
        // to 0
        if (Y != 0)
        {
             
            // Insert Y
            // into PQ
            PQ.add(Y);
        }
         
        // Update cntOp
        cntOp += 1;
    }
    return cntOp;
}
 
// Driver code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int N = arr.length;
 
    System.out.print(cntMaxOperationToMakeN_1_0(arr, N));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python3 program to implement
# the above approach
 
# Function to count maximum number of steps
# to make (N - 1) array elements to 0
def cntMaxOperationToMakeN_1_0(arr, N):
 
    # Stores maximum count of steps to make
    # (N - 1) elements equal to 0
    cntOp = 0
 
    # Stores array elements
    PQ = []
 
    # Traverse the array
    for i in range(N):
 
        # Insert arr[i] into PQ
        PQ.append(arr[i])
    PQ = sorted(PQ)
 
    # Extract top 2 elements from the array
    # while (N - 1) array elements become 0
    while (len(PQ) > 1):
 
        # Stores top element
        # of PQ
        X = PQ[-1]
 
        # Pop the top element
        # of PQ.
        del PQ[-1]
 
        # Stores top element
        # of PQ
        Y = PQ[-1]
 
        # Pop the top element
        # of PQ.
        del PQ[-1]
 
        # Update X
        X -= 1
 
        # Update Y
        Y -= 1
 
        # If X is not equal to 0
        if (X != 0):
 
            # Insert X into PQ
            PQ.append(X)
 
        # if Y is not equal
        # to 0
        if (Y != 0):
 
            # Insert Y
            # into PQ
            PQ.append(Y)
 
        # Update cntOp
        cntOp += 1
        PQ = sorted(PQ)
    return cntOp
 
# Driver Code
if __name__ == '__main__':
 
    arr = [1, 2, 3, 4, 5]
    N = len(arr)
    print (cntMaxOperationToMakeN_1_0(arr, N))
 
    # This code is contributed by mohit kumar 29.

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to count maximum number of steps
  // to make (N - 1) array elements to 0
  static int cntMaxOperationToMakeN_1_0(int[] arr, int N)
  {
 
    // Stores maximum count of steps to make
    // (N - 1) elements equal to 0
    int cntOp = 0;
 
    // Stores array elements
    List<int> PQ = new List<int>();
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Insert arr[i] into PQ
      PQ.Add(arr[i]);
    }
 
    PQ.Sort();
    PQ.Reverse();
 
    // Extract top 2 elements from the array
    // while (N - 1) array elements become 0
    while (PQ.Count > 1) {
 
      // Stores top element
      // of PQ
      int X = PQ[0];
 
      // Pop the top element
      // of PQ.
      PQ.RemoveAt(0);
 
      // Stores top element
      // of PQ
      int Y = PQ[0];
 
      // Pop the top element
      // of PQ.
      PQ.RemoveAt(0);
 
      // Update X
      X--;
 
      // Update Y
      Y--;
 
      // If X is not equal to 0
      if (X != 0) {
 
        // Insert X into PQ
        PQ.Add(X);
        PQ.Sort();
        PQ.Reverse();
      }
 
      // if Y is not equal
      // to 0
      if (Y != 0) {
 
        // Insert Y
        // into PQ
        PQ.Add(Y);
        PQ.Sort();
        PQ.Reverse();
      }
 
      // Update cntOp
      cntOp += 1;
    }
 
    return cntOp;
  }
 
  // Driver code
  static void Main() {
    int[] arr = { 1, 2, 3, 4, 5 };
 
    int N = arr.Length;
 
    Console.WriteLine(cntMaxOperationToMakeN_1_0(arr, N));
  }
}
 
// This code is contributed by divyesh072019
Output: 
7

 

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

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