Maximize element at index K in an array with at most sum M and difference between adjacent elements at most 1

Given a positive integer N, the task is to construct an array of length N and find the maximum value at index K such that the sum of all the array elements is at most M and the absolute difference between any two consecutive array elements is at most 1.

Examples:

Input: N = 3, M = 7, K = 1
Output: 3
Explanation: 
According to the given constraints, the array with values {2, 3, 2}maximizes the value at index 1. Therefore, the required output is 3.

Input: N = 3, M = 8, K = 1
Output: 3

Approach: The idea is to achieve the maximum value at index K and to decrease the sum of other elements to meet the criteria of the sum of the array to be at most M. Follow the steps below:

• Let the value at index K be X. So the element at K - 1 is X - 1, at K - 2 is X - 2 and so on.
• At index 0 the value is X - K. Similarly, at index K + 1 the value is X - 1 and so on upto X - (N - K - 1) at index N - 1.
• So to achieve the maximum value at index K, the array structure would be X - K, X - (K - 1), ...., X - 2, X - 1, X, X - 1, X - 2, ....., X - (N - K - 1).
• So after arranging the equation, it becomes K * X - (1 + 2 + .... + K) + X + (N - K - 1) * X - (1 + 2 + .... + (N - K - 1)) ? M.

Follow the steps to solve the above equation:

• Calculate (1 + 2 + .... + K) using K * (K + 1) / 2 and (1 + 2 + ..... + (N - K - 1)) using (N - K - 1) * (N - K) / 2 and store in S1 and S2 respectively.
• This reduces the equation to X * (K + 1 + N - K - 1) ? M + S1 + S2.
• Now, the maximum value of X can be obtained by calculating (M + S1 + S2) / N.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the maximum
// possible value at index K
void maxValueAtIndexK(int N, int K, int M)
{
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;

    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;

    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;

    // Print the answer
    cout << X;
}

// Driver Code
int main()
{
    // Given N, K & M
    int N = 3, K = 1, M = 7;
    maxValueAtIndexK(N, K, M);

    return 0;
}

// Java program for the above approach
import java.io.*;

class GFG{

// Function to calculate the maximum
// possible value at index K
static void maxValueAtIndexK(int N, int K,
                             int M)
{
    
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;

    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;

    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;

    // Print the answer
    System.out.println(X);
}

// Driver Code
public static void main(String[] args)
{
    
    // Given N, K & M
    int N = 3, K = 1, M = 7;
    
    maxValueAtIndexK(N, K, M);
}
}

// This code is contributed by Dharanendra L V

# Python program for the above approach

# Function to calculate the maximum
# possible value at index K
def maxValueAtIndexK(N, K, M):

    # Stores the sum of elements
    # in the left and right of index K
    S1 = 0; S2 = 0;
    S1 = K * (K + 1) // 2;
    S2 = (N - K - 1) * (N - K) // 2;

    # Stores the maximum
    # possible value at index K
    X = (M + S1 + S2) // N;

    # Print the answer
    print(X);

# Driver Code
if __name__ == '__main__':

    # Given N, K & M
    N = 3; K = 1; M = 7;
    maxValueAtIndexK(N, K, M);

# This code is contributed by 29AjayKumar 

// C# program for the above approach
using System;

class GFG{

// Function to calculate the maximum
// possible value at index K
static void maxValueAtIndexK(int N, int K,
                             int M)
{
    
    // Stores the sum of elements
    // in the left and right of index K
    int S1 = 0, S2 = 0;

    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;

    // Stores the maximum
    // possible value at index K
    int X = (M + S1 + S2) / N;

    // Print the answer
    Console.WriteLine(X);
}

// Driver Code
static public void Main()
{
    
    // Given N, K & M
    int N = 3, K = 1, M = 7;
    
    maxValueAtIndexK(N, K, M);
}
}

// This code is contributed by Dharanendra L V

<script>
// JavaScript program for
// the above approach
 
// Function to calculate the maximum
// possible value at index K
function maxValueAtIndexK(N, K, M)
{
    // Stores the sum of elements
    // in the left and right of index K
    let S1 = 0, S2 = 0;

    S1 = K * (K + 1) / 2;
    S2 = (N - K - 1) * (N - K) / 2;

    // Stores the maximum
    // possible value at index K
    let X = (M + S1 + S2) / N;

    // Print the answer
     document.write(X);
}

// Driver Code
 
    // Given N, K & M
    let N = 3, K = 1, M = 7;
    maxValueAtIndexK(N, K, M);
 
</script>

3

Time Complexity: O(1)
Auxiliary Space: O(1)