Maximize the minimum array element by M subarray increments of size S

Given an array arr[] of N integers and two integers S and M, the task is to maximize the minimum array element by incrementing any subarray of size S by 1, M number of times.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}, S = 2, M = 3
Output: 3
Explanation:
Below are the operations performed:
Operation 1: Select subarray {1, 2} and after increment, array arr[] becomes = {2, 3, 3, 4, 5, 6}.
Operation 2: Select subarray {2, 3} and after increment, array arr[] becomes = {3, 4, 3, 4, 5, 6}.
Operation 3: Select subarray {3, 4} and after increment, array arr[] becomes = {4, 5, 3, 4, 5, 6}.
After the above operations, the minimum element of the array is 3.

Input: arr[] = {3, 5, 2, 7, 3}, S = 3, M = 3
Output: 4
Explanation:
Below are the operations performed:
Operation 1: Select subarray {3, 5, 2} and after increment, array arr[] becomes = {4, 6, 3, 7, 3}.
Operation 2: Select subarray {4, 6, 3} and after increment, array arr[] becomes = {5, 7, 4, 7, 3}.
Operation 3: Select subarray {4, 7, 3} and after increment, array arr[] becomes = {5, 7, 5, 8, 4}.
After the above operations, the minimum element of the array is 4.

Approach: The idea is to find the minimum element of the array M number of times and increment subarrays of size S from that minimum element by 1. Follow the steps below to solve the problem:

• Traverse over the array M number of times and for each iteration do the following:
  • Find the minimum element of the array arr[]. Let the first index of the minimum element be idx.
  • Increment the current minimum element by 1.
  • Now take two pointers leftIdx as idx – 1 and rightIdx as idx + 1.
  • If the element at leftIdx is less than the element at rightIdx then increment A[leftIndex] by 1 and decrement leftIndex by 1. Otherwise, increment A[rightIndex] by 1 and rightIndex by 1. Continue this step till (S – 1) elements are processed.
• After the above iterations, print the minimum element of the updated array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return index of minimum
// element in the array
int min(int a[], int n)
{
    // Initialize a[0] as minValue
    int minIndex = 0, minValue = a[0], i;

    // Traverse the array
    for (i = 1; i < n; i++) {

        // If a[i] < existing minValue
        if (a[i] < minValue) {
            minValue = a[i];
            minIndex = i;
        }
    }

    // Return the minimum index
    return minIndex;
}

// Function that maximize the minimum
// element of array after incrementing
// subarray of size S by 1, M times
int maximizeMin(int A[], int N,
                int S, int M)
{
    int minIndex, left, right, i, j;

    // Iterating through the array
    // for M times
    for (i = 0; i < M; i++) {

        // Find minimum element index
        minIndex = min(A, N);

        // Increment the minimum value
        A[minIndex]++;

        // Storing the left index
        // and right index
        left = minIndex - 1;
        right = minIndex + 1;

        // Incrementing S - 1 minimum
        // elements to the left and
        // right of minValue
        for (j = 0; j < S - 1; j++) {

            // Reached extreme left
            if (left == -1)
                A[right++]++;

            // Reached extreme right
            else if (right == N)
                A[left--]++;

            else {

                // Left value is minimum
                if (A[left] < A[right])
                    A[left--]++;

                // Right value is minimum
                else
                    A[right++]++;
            }
        }
    }

    // Find the minValue in A[] after
    // M operations
    minIndex = min(A, N);

    // Return the minimum value
    return A[minIndex];
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int S = 2, M = 3;

    // Function Call
    cout << maximizeMin(arr, N, S, M);

    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class solution{

// Function to return index
// of minimum element in the
// array
static int min1(int a[], int n)
{
  // Initialize a[0] as
  // minValue
  int minIndex = 0,
      minValue = a[0], i;

  // Traverse the array
  for (i = 1; i < n; i++)
  {
    // If a[i] < existing
    // minValue
    if (a[i] < minValue)
    {
      minValue = a[i];
      minIndex = i;
    }
  }

  // Return the minimum index
  return minIndex;
}

// Function that maximize the minimum
// element of array after incrementing
// subarray of size S by 1, M times
static int maximizeMin(int A[], int N,
                       int S, int M)
{
  int minIndex, left, right, i, j;

  // Iterating through the
  // array or M times
  for (i = 0; i < M; i++)
  {
    // Find minimum element
    // index
    minIndex = min1(A, N);

    // Increment the minimum
    // value
    A[minIndex]++;

    // Storing the left index
    // and right index
    left = minIndex - 1;
    right = minIndex + 1;

    // Incrementing S - 1 minimum
    // elements to the left and
    // right of minValue
    for (j = 0; j < S - 1; j++)
    {
      // Reached extreme left
      if (left == -1)
        A[right++]++;

      // Reached extreme right
      else if (right == N)
        A[left--]++;

      else
      {
        // Left value is minimum
        if (A[left] < A[right])
          A[left--]++;

        // Right value is minimum
        else
          A[right++]++;
      }
    }
  }

  // Find the minValue in A[] after
  // M operations
  minIndex = min1(A, N);

  // Return the minimum value
  return A[minIndex];
}

// Driver Code
public static void main(String args[])
{
  int []arr = {1, 2, 3,
               4, 5, 6};
  int N = arr.length;
  int S = 2, M = 3;

  // Function Call
  System.out.print(maximizeMin(arr, N, S, M));
}
}

// This code is contributed by SURENDRA_GANGWAR

Python3

# Python3 program for the above approach

# Function to return index of minimum
# element in the array
def min(a, n):

    # Initialize a[0] as minValue
    minIndex = 0
    minValue = a[0]

    # Traverse the array
    for i in range(1, n):

        # If a[i] < existing minValue
        if (a[i] < minValue):
            minValue = a[i]
            minIndex = i

    # Return the minimum index
    return minIndex

# Function that maximize the minimum
# element of array after incrementing
# subarray of size S by 1, M times
def maximizeMin(A, N, S, M):

    minIndex, left, right = 0, 0, 0

    # Iterating through the array
    # for M times
    for i in range(M):

        # Find minimum element index
        minIndex = min(A, N)

        # Increment the minimum value
        A[minIndex] += 1

        # Storing the left index
        # and right index
        left = minIndex - 1
        right = minIndex + 1

        # Incrementing S - 1 minimum
        # elements to the left and
        # right of minValue
        for j in range(S - 1):

            # Reached extreme left
            if (left == -1):
                A[right] += 1
                right += 1

            # Reached extreme right
            elif (right == N):
                A[left] += 1
                left -= 1

            else:

                # Left value is minimum
                if (A[left] < A[right]):
                    A[left] += 1
                    left -= 1

                # Right value is minimum
                else:
                    A[right] += 1
                    right += 1

    # Find the minValue in A[] after
    # M operations
    minIndex = min(A, N)

    # Return the minimum value
    return A[minIndex]

# Driver Code
if __name__ == '__main__':

    arr = [ 1, 2, 3, 4, 5, 6 ]
    N = len(arr)
    S = 2
    M = 3

    #Function Call
    print(maximizeMin(arr, N, S, M))

# This code is contributed by mohit kumar 29

C#

// C# program for the
// above approach
using System;
class GFG{

// Function to return index
// of minimum element in the
// array
static int min1(int[] a,
                int n)
{
  // Initialize a[0] as
  // minValue
  int minIndex = 0,
      minValue = a[0], i;

  // Traverse the array
  for (i = 1; i < n; i++)
  {
    // If a[i] < existing
    // minValue
    if (a[i] < minValue)
    {
      minValue = a[i];
      minIndex = i;
    }
  }

  // Return the minimum
  // index
  return minIndex;
}

// Function that maximize the
// minimum element of array
// after incrementing subarray
// of size S by 1, M times
static int maximizeMin(int[] A, int N,
                       int S, int M)
{
  int minIndex, left, right, i, j;

  // Iterating through the
  // array or M times
  for (i = 0; i < M; i++)
  {
    // Find minimum element
    // index
    minIndex = min1(A, N);

    // Increment the minimum
    // value
    A[minIndex]++;

    // Storing the left index
    // and right index
    left = minIndex - 1;
    right = minIndex + 1;

    // Incrementing S - 1 minimum
    // elements to the left and
    // right of minValue
    for (j = 0; j < S - 1; j++)
    {
      // Reached extreme left
      if (left == -1)
        A[right++]++;

      // Reached extreme right
      else if (right == N)
        A[left--]++;

      else
      {
        // Left value is minimum
        if (A[left] < A[right])
          A[left--]++;

        // Right value is minimum
        else
          A[right++]++;
      }
    }
  }

  // Find the minValue in A[] after
  // M operations
  minIndex = min1(A, N);

  // Return the minimum value
  return A[minIndex];
}

// Driver code
static void Main()
{
  int[] arr = {1, 2, 3,
               4, 5, 6};
  int N = arr.Length;
  int S = 2, M = 3;

  // Function Call
  Console.Write(maximizeMin(arr, N,
                            S, M));
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to return index
// of minimum element in the
// array
function min1(a, n)
{
  // Initialize a[0] as
  // minValue
  let minIndex = 0,
      minValue = a[0], i;

  // Traverse the array
  for (i = 1; i < n; i++)
  {
    // If a[i] < existing
    // minValue
    if (a[i] < minValue)
    {
      minValue = a[i];
      minIndex = i;
    }
  }

  // Return the minimum index
  return minIndex;
}

// Function that maximize the minimum
// element of array after incrementing
// subarray of size S by 1, M times
function maximizeMin(A, N, S, M)
{
  let minIndex, left, right, i, j;

  // Iterating through the
  // array or M times
  for (i = 0; i < M; i++)
  {
    // Find minimum element
    // index
    minIndex = min1(A, N);

    // Increment the minimum
    // value
    A[minIndex]++;

    // Storing the left index
    // and right index
    left = minIndex - 1;
    right = minIndex + 1;

    // Incrementing S - 1 minimum
    // elements to the left and
    // right of minValue
    for (j = 0; j < S - 1; j++)
    {
      // Reached extreme left
      if (left == -1)
        A[right++]++;

      // Reached extreme right
      else if (right == N)
        A[left--]++;

      else
      {
        // Left value is minimum
        if (A[left] < A[right])
          A[left--]++;

        // Right value is minimum
        else
          A[right++]++;
      }
    }
  }

  // Find the minValue in A[] after
  // M operations
  minIndex = min1(A, N);

  // Return the minimum value
  return A[minIndex];
}


// Driver Code

    let arr = [1, 2, 3,
               4, 5, 6];
  let N = arr.length;
  let S = 2, M = 3;

  // Function Call
  document.write(maximizeMin(arr, N, S, M));

</script>

Output:

3

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