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

  • Difficulty Level : Hard
  • Last Updated : 15 Mar, 2021

Given a positive integer N, the task is to construct an array of length N and find the maximize the 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++




// 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




// 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

Python3




# 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;
 
    # Prthe 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#




// 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

Javascript




<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>
Output: 
3

 

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

 

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