Maximize pair decrements required to reduce all array elements except one to 0

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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

Javascript

<script>
    // Javascript program to implement the above approach
     
    // Function to count maximum number of steps
    // to make (N - 1) array elements to 0
    function cntMaxOperationToMakeN_1_0(arr, N)
    {
 
      // Stores maximum count of steps to make
      // (N - 1) elements equal to 0
      let cntOp = 0;
 
      // Stores array elements
      let PQ = [];
 
      // Traverse the array
      for (let i = 0; i < N; i++) {
 
        // Insert arr[i] into PQ
        PQ.push(arr[i]);
      }
 
      PQ.sort(function(a, b){return a - b});
      PQ.reverse();
 
      // Extract top 2 elements from the array
      // while (N - 1) array elements become 0
      while (PQ.length > 1) {
 
        // Stores top element
        // of PQ
        let X = PQ[0];
 
        // Pop the top element
        // of PQ.
        PQ.shift();
 
        // Stores top element
        // of PQ
        let Y = PQ[0];
 
        // Pop the top element
        // of PQ.
        PQ.shift();
 
        // Update X
        X--;
 
        // Update Y
        Y--;
 
        // If X is not equal to 0
        if (X != 0) {
 
          // Insert X into PQ
          PQ.push(X);
          PQ.sort(function(a, b){return a - b});
          PQ.reverse();
        }
 
        // if Y is not equal
        // to 0
        if (Y != 0) {
 
          // Insert Y
          // into PQ
          PQ.push(Y);
          PQ.sort(function(a, b){return a - b});
          PQ.reverse();
        }
 
        // Update cntOp
        cntOp += 1;
      }
 
      return cntOp;
    }
     
    let arr = [ 1, 2, 3, 4, 5 ];
    let N = arr.length;
    document.write(cntMaxOperationToMakeN_1_0(arr, N));
     
    // This code is contributed by mukesh07.
</script>

Output

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

Approach: Using the Two Pointers approach

Algorithm steps:

Sort the array arr in descending order using std::sort and the greater<int> comparator.
Initialize two pointers, left and right, where left starts at index 0 and right starts at index N – 2 (since we need to make (N – 1) elements equal to 0).
Initialize a variable cntOp to keep track of the count of steps.
Iterate until the left and right pointers meet or cross each other.
In each iteration:
- Subtract 1 from both elements pointed to by the left and right pointers.
- If either element becomes non-positive:
- Move the left pointer to the right (increment it) if the left element becomes non-positive.
- Move the right pointer to the left (decrement it) if the right element becomes non-positive.
- Increment the count of steps cntOp by 1.
After the iteration is complete, return the value of cntOp, which represents the maximum number of steps required to make (N – 1) elements equal to 0.

Below is the implementation of the above approach:

C++

//C++ code for the above approach
#include <iostream>
#include <algorithm>
using namespace std;
 
// Function to count maximum number of steps
// to make (N - 1) array elements equal to 0
int cntMaxOperationToMakeN_1_0(int arr[], int N)
{
    // Sort the array in descending order
    sort(arr, arr + N, greater<int>());
 
    // Initialize two pointers
    int left = 0;
    int right = N - 2; // (N - 1) elements to make 0
 
    // Count the number of steps
    int cntOp = 0;
 
    // Perform steps until left and right pointers meet
    while (left <= right) {
        // Subtract 1 from both elements
        arr[left]--;
        arr[right]--;
 
        // If either element becomes non-positive,
        // move the pointer accordingly
        if (arr[left] <= 0)
            left++;
        if (arr[right] <= 0)
            right--;
 
        // Increment the count of steps
        cntOp++;
    }
 
    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

import java.util.Arrays;
import java.util.Collections;
 
public class Main {
 
    // Function to count maximum number of steps
    // to make (N - 1) array elements equal to 0
    static int cntMaxOperationToMakeN_1_0(int arr[], int N) {
        // Sort the array in descending order
        Arrays.sort(arr);
        reverseArray(arr);
 
        // Initialize two pointers
        int left = 0;
        int right = N - 2; // (N - 1) elements to make 0
 
        // Count the number of steps
        int cntOp = 0;
 
        // Perform steps until left and right pointers meet
        while (left <= right) {
            // Subtract 1 from both elements
            arr[left]--;
            arr[right]--;
 
            // If either element becomes non-positive,
            // move the pointer accordingly
            if (arr[left] <= 0)
                left++;
            if (arr[right] <= 0)
                right--;
 
            // Increment the count of steps
            cntOp++;
        }
 
        return cntOp;
    }
 
    // Function to reverse the array
    static void reverseArray(int arr[]) {
        int left = 0;
        int right = arr.length - 1;
        while (left < right) {
            int temp = arr[left];
            arr[left] = arr[right];
            arr[right] = temp;
            left++;
            right--;
        }
    }
 
    // Driver Code
    public static void main(String[] args) {
        int arr[] = {1, 2, 3, 4, 5};
        int N = arr.length;
 
        System.out.println(cntMaxOperationToMakeN_1_0(arr, N));
    }
}

Python3

def cntMaxOperationToMakeN_1_0(arr, N):
    # Sort the array in descending order
    arr.sort(reverse=True)
 
    # Initialize two pointers
    left = 0
    right = N - 2  # (N - 1) elements to make 0
 
    # Count the number of steps
    cntOp = 0
 
    # Perform steps until left and right pointers meet
    while left <= right:
        # Subtract 1 from both elements
        arr[left] -= 1
        arr[right] -= 1
 
        # If either element becomes non-positive,
        # move the pointer accordingly
        if arr[left] <= 0:
            left += 1
        if arr[right] <= 0:
            right -= 1
 
        # Increment the count of steps
        cntOp += 1
 
    return cntOp
 
# Driver Code
arr = [1, 2, 3, 4, 5]
N = len(arr)
print(cntMaxOperationToMakeN_1_0(arr, N))

C#

using System;
using System.Linq;
 
namespace MaxOperationsToMakeZero
{
    class Program
    {
        static int CountMaxOperationToMakeN_1_0(int[] arr, int N)
        {
            // Sort the array in descending order
            Array.Sort(arr, (x, y) => y.CompareTo(x));
 
            // Initialize two pointers
            int left = 0;
            int right = N - 2; // (N - 1) elements to make 0
 
            // Count the number of steps
            int cntOp = 0;
 
            // Perform steps until left and right pointers meet
            while (left <= right)
            {
                // Subtract 1 from both elements
                arr[left]--;
                arr[right]--;
 
                // If either element becomes non-positive,
                // move the pointer accordingly
                if (arr[left] <= 0)
                    left++;
                if (arr[right] <= 0)
                    right--;
 
                // Increment the count of steps
                cntOp++;
            }
 
            return cntOp;
        }
 
        static void Main(string[] args)
        {
            int[] arr = { 1, 2, 3, 4, 5 };
            int N = arr.Length;
 
            Console.WriteLine(CountMaxOperationToMakeN_1_0(arr, N));
        }
    }
}

Javascript

// Function to count maximum number of steps
// to make (N - 1) array elements equal to 0
function cntMaxOperationToMakeN_1_0(arr, N) {
    // Sort the array in descending order
    arr.sort((a, b) => b - a);
 
    // Initialize two pointers
    let left = 0;
    let right = N - 2; // (N - 1) elements to make 0
 
    // Count the number of steps
    let cntOp = 0;
 
    // Perform steps until left and right pointers meet
    while (left <= right) {
        // Subtract 1 from both elements
        arr[left]--;
        arr[right]--;
 
        // If either element becomes non-positive,
        // move the pointer accordingly
        if (arr[left] <= 0)
            left++;
        if (arr[right] <= 0)
            right--;
 
        // Increment the count of steps
        cntOp++;
    }
 
    return cntOp;
}
 
// Driver Code
const arr = [1, 2, 3, 4, 5];
const N = arr.length;
 
console.log(cntMaxOperationToMakeN_1_0(arr, N));

Output

Time Complexity: O(N log N) , because of the sorting operation.
Auxiliary Space: O(1) because the algorithm only uses a constant amount of additional space for the pointers and variables.

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