Given an array A[] consisting of N integers(1-based indexing), the task is to find the minimum value of the function ![\sum_{i = 1}^{N}(A[i] - (X + i))](https://cdn.geeksforgeeks.org/static/gallery/images/logo.png "Rendered by QuickLaTeX.com"){kind=logo}
for any possible value of X.
Examples:
Input: A[] = {1, 2, 3, 4}
Output: 0
Explanation:
Consider the value of X as 0, then the value of the given function is (1 – 1 + 2 – 2 + 3 – 3 + 4 – 4) = 0, which is minimum.Input: A[] = {5, 3, 9}
Output: 5
Approach: The given problem can be solved based on the following observations:
- Consider a function as (B[i] = A[i] ? i), then to minimize the value of
![\sum_{i = 1}^{N}(B[i] - X)](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20181%2033'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
, the idea is to choose the value of X as the median of the array B[] such that the sum is minimized.
Follow the steps to solve the problem:
- Initialize an array, say B[] that stores the value of (A[i] – i) for every possible value of the array A[].
- Traverse the given array A[] and for each index i, update the value of B[i] as (A[i] – i).
- Sort the array B[] in ascending order and find the value X as the median of the array B[].
- After completing the above steps, find the value of the given function
![\sum_{i = 1}^{N}(A[i] - (X + i))](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20241%2033'%3E%3C/svg%3E "Rendered by QuickLaTeX.com")
for the calculated value of X.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find minimum value of// the given functionint minimizeFunction(int A[], int N){ // Stores the value of A[i] - i int B[N]; // Traverse the given array A[] for (int i = 0; i < N; i++) { // Update the value of B[i] B[i] = A[i] - i - 1; } // Sort the array B[] sort(B, B + N); // Calculate the median of the // array B[] int median = (B[int(floor((N - 1) / 2.0))] + B[int(ceil((N - 1) / 2.0))]) / 2; // Stores the minimum value of // the function int minValue = 0; for (int i = 0; i < N; i++) { // Update the minValue minValue += abs(A[i] - (median + i + 1)); } // Return the minimum value return minValue;}// Driver Codeint main(){ int A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int N = sizeof(A) / sizeof(A[0]); cout << minimizeFunction(A, N); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.lang.Math;import java.util.*;class GFG { public static int minimizeFunction(int A[], int N) { // Stores the value of A[i] - i int B[] = new int[N]; // Traverse the given array A[] for (int i = 0; i < N; i++) { // Update the value of B[i] B[i] = A[i] - i - 1; } // Sort the array B[] Arrays.sort(B); // Calculate the median of the // array B[] int median = (B[(int)(Math.floor((N - 1) / 2.0))] + B[(int)(Math.ceil((N - 1) / 2.0))]) / 2; // Stores the minimum value of // the function int minValue = 0; for (int i = 0; i < N; i++) { // Update the minValue minValue += Math.abs(A[i] - (median + i + 1)); } // Return the minimum value return minValue; } public static void main(String[] args) { int A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int N = A.length; System.out.println(minimizeFunction(A, N)); }}// This code is contributed by sam_2200. |
Python3
# Python3 program for the above approachfrom math import floor, ceil# Function to find minimum value of# the given functiondef minimizeFunction(A, N): # Stores the value of A[i] - i B = [0] * N # Traverse the given array A[] for i in range(N): # Update the value of B[i] B[i] = A[i] - i - 1 # Sort the array B[] B = sorted(B) # Calculate the median of the # array B[] x, y = int(floor((N - 1) / 2.0)), int(ceil((N - 1) / 2.0)) median = (B[x] + B[y]) / 2 # Stores the minimum value of # the function minValue = 0 for i in range(N): # Update the minValue minValue += abs(A[i] - (median + i + 1)) # Return the minimum value return int(minValue)# Driver Codeif __name__ == '__main__': A = [1, 2, 3, 4, 5, 6, 7, 8, 9] N = len(A) print(minimizeFunction(A, N))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG { public static int minimizeFunction(int[] A, int N) { // Stores the value of A[i] - i int[] B = new int[N]; // Traverse the given array A[] for (int i = 0; i < N; i++) { // Update the value of B[i] B[i] = A[i] - i - 1; } // Sort the array B[] Array.Sort(B); // Calculate the median of the // array B[] int median = (B[(int)(Math.Floor((N - 1) / 2.0))] + B[(int)(Math.Ceiling((N - 1) / 2.0))]) / 2; // Stores the minimum value of // the function int minValue = 0; for (int i = 0; i < N; i++) { // Update the minValue minValue += Math.Abs(A[i] - (median + i + 1)); } // Return the minimum value return minValue; } static void Main() { int []A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; int N = A.Length; Console.WriteLine(minimizeFunction(A, N)); }}// This code is contributed by SoumikMondal. |
Javascript
<script>// Javascript program for the above approachfunction minimizeFunction(A, N){ // Stores the value of A[i] - i let B = Array.from({length: N}, (_, i) => 0); // Traverse the given array A[] for (let i = 0; i < N; i++) { // Update the value of B[i] B[i] = A[i] - i - 1; } // Sort the array B[] B.sort(); // Calculate the median of the // array B[] let median = (B[(Math.floor((N - 1) / 2.0))] + B[(Math.ceil((N - 1) / 2.0))]) / 2; // Stores the minimum value of // the function let minValue = 0; for (let i = 0; i < N; i++) { // Update the minValue minValue += Math.abs(A[i] - (median + i + 1)); } // Return the minimum value return minValue; }// Driver Code let A = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]; let N = A.length; document.write(minimizeFunction(A, N));</script> |
Output:
0
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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