Minimum value of maximum absolute difference of all adjacent pairs in an Array

Given an array arr, containing non-negative integers and (-1)s, of size N, the task is to replace those (-1)s with a common non-negative integer such that the maximum absolute difference of all adjacent pairs is minimum. Print this minimum possible value of the maximum absolute difference.
Examples:
Input: arr = {-1, -1, 11, -1, 3, -1}
Output: 4
Replace every -1 element with 7. Now the maximum absolute difference of all adjacent pairs is minimum which is equal to 4Input: arr = {4, -1}
Output: 0
Approach:
- Consider only those non-missing elements that are adjacent to at least one missing element.
- Find the maximum element and the minimum element among them.
- We need to find a value that minimizes the maximum absolute difference between the common value and these values.
- The optimal value is equals to
(minimum element + maximum element) / 2
Below is the implementation of the above approach:
C++
// C++ program to find the minimum value// of maximum absolute difference of// all adjacent pairs in an Array#include <bits/stdc++.h>using namespace std;// Function to find the minimum possible// value of the maximum absolute difference.int maximumAbsolute(int arr[], int n){ // To store minimum and maximum elements int mn = INT_MAX; int mx = INT_MIN; for (int i = 0; i < n; i++) { // If right side element is equals -1 // and left side is not equals -1 if (i > 0 && arr[i] == -1 && arr[i - 1] != -1) { mn = min(mn, arr[i - 1]); mx = max(mx, arr[i - 1]); } // If left side element is equals -1 // and right side is not equals -1 if (i < n - 1 && arr[i] == -1 && arr[i + 1] != -1) { mn = min(mn, arr[i + 1]); mx = max(mx, arr[i + 1]); } } // Calculating the common integer // which needs to be replaced with int common_integer = (mn + mx) / 2; // Replace all -1 elements // with the common integer for (int i = 0; i < n; i++) { if (arr[i] == -1) arr[i] = common_integer; } int max_diff = 0; // Calculating the maximum // absolute difference for (int i = 0; i < n - 1; i++) { int diff = abs(arr[i] - arr[i + 1]); if (diff > max_diff) max_diff = diff; } // Return the maximum absolute difference return max_diff;}// Driver Codeint main(){ int arr[] = { -1, -1, 11, -1, 3, -1 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << maximumAbsolute(arr, n); return 0;} |
Java
// Java program to find the minimum value// of maximum absolute difference of// all adjacent pairs in an Arrayimport java.util.*;class GFG{ // Function to find the minimum possible// value of the maximum absolute difference.static int maximumAbsolute(int arr[], int n){ // To store minimum and maximum elements int mn = Integer.MAX_VALUE; int mx = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { // If right side element is equals -1 // and left side is not equals -1 if (i > 0 && arr[i] == -1 && arr[i - 1] != -1) { mn = Math.min(mn, arr[i - 1]); mx = Math.max(mx, arr[i - 1]); } // If left side element is equals -1 // and right side is not equals -1 if (i < n - 1 && arr[i] == -1 && arr[i + 1] != -1) { mn = Math.min(mn, arr[i + 1]); mx = Math.max(mx, arr[i + 1]); } } // Calculating the common integer // which needs to be replaced with int common_integer = (mn + mx) / 2; // Replace all -1 elements // with the common integer for (int i = 0; i < n; i++) { if (arr[i] == -1) arr[i] = common_integer; } int max_diff = 0; // Calculating the maximum // absolute difference for (int i = 0; i < n - 1; i++) { int diff = Math.abs(arr[i] - arr[i + 1]); if (diff > max_diff) max_diff = diff; } // Return the maximum absolute difference return max_diff;} // Driver Codepublic static void main(String[] args){ int arr[] = { -1, -1, 11, -1, 3, -1 }; int n = arr.length; // Function call System.out.print(maximumAbsolute(arr, n));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to find the minimum value# of maximum absolute difference of# all adjacent pairs in an Array# Function to find the minimum possible# value of the maximum absolute difference.def maximumAbsolute(arr, n): # To store minimum and maximum elements mn = 10**9 mx = -10**9 for i in range(n): # If right side element is equals -1 # and left side is not equals -1 if (i > 0 and arr[i] == -1 and arr[i - 1] != -1): mn = min(mn, arr[i - 1]) mx = max(mx, arr[i - 1]) # If left side element is equals -1 # and right side is not equals -1 if (i < n - 1 and arr[i] == -1 and arr[i + 1] != -1): mn = min(mn, arr[i + 1]) mx = max(mx, arr[i + 1]) # Calculating the common integer # which needs to be replaced with common_integer = (mn + mx) // 2 # Replace all -1 elements # with the common integer for i in range(n): if (arr[i] == -1): arr[i] = common_integer max_diff = 0 # Calculating the maximum # absolute difference for i in range(n-1): diff = abs(arr[i] - arr[i + 1]) if (diff > max_diff): max_diff = diff # Return the maximum absolute difference return max_diff# Driver Codeif __name__ == '__main__': arr=[-1, -1, 11, -1, 3, -1] n = len(arr) # Function call print(maximumAbsolute(arr, n))# This code is contributed by mohit kumar 29 |
C#
// C# program to find the minimum value// of maximum absolute difference of// all adjacent pairs in an Arrayusing System;class GFG{ // Function to find the minimum possible // value of the maximum absolute difference. static int maximumAbsolute(int []arr, int n) { // To store minimum and maximum elements int mn = int.MaxValue; int mx = int.MinValue; for (int i = 0; i < n; i++) { // If right side element is equals -1 // and left side is not equals -1 if (i > 0 && arr[i] == -1 && arr[i - 1] != -1) { mn = Math.Min(mn, arr[i - 1]); mx = Math.Max(mx, arr[i - 1]); } // If left side element is equals -1 // and right side is not equals -1 if (i < n - 1 && arr[i] == -1 && arr[i + 1] != -1) { mn = Math.Min(mn, arr[i + 1]); mx = Math.Max(mx, arr[i + 1]); } } // Calculating the common integer // which needs to be replaced with int common_integer = (mn + mx) / 2; // Replace all -1 elements // with the common integer for (int i = 0; i < n; i++) { if (arr[i] == -1) arr[i] = common_integer; } int max_diff = 0; // Calculating the maximum // absolute difference for (int i = 0; i < n - 1; i++) { int diff = Math.Abs(arr[i] - arr[i + 1]); if (diff > max_diff) max_diff = diff; } // Return the maximum absolute difference return max_diff; } // Driver Code public static void Main(string[] args) { int []arr = { -1, -1, 11, -1, 3, -1 }; int n = arr.Length; // Function call Console.Write(maximumAbsolute(arr, n)); }}// This code is contributed by Yash_R |
Javascript
<script>// Javascript program to find the minimum value// of maximum absolute difference of// all adjacent pairs in an Array// Function to find the minimum possible// value of the maximum absolute difference.function maximumAbsolute(arr, n) { // To store minimum and maximum elements var mn = Number.MAX_VALUE; var mx = Number.MIN_VALUE; for(i = 0; i < n; i++) { // If right side element is equals -1 // and left side is not equals -1 if (i > 0 && arr[i] == -1 && arr[i - 1] != -1) { mn = Math.min(mn, arr[i - 1]); mx = Math.max(mx, arr[i - 1]); } // If left side element is equals -1 // and right side is not equals -1 if (i < n - 1 && arr[i] == -1 && arr[i + 1] != -1) { mn = Math.min(mn, arr[i + 1]); mx = Math.max(mx, arr[i + 1]); } } // Calculating the common integer // which needs to be replaced with var common_integer = (mn + mx) / 2; // Replace all -1 elements // with the common integer for(i = 0; i < n; i++) { if (arr[i] == -1) arr[i] = common_integer; } var max_diff = 0; // Calculating the maximum // absolute difference for(i = 0; i < n - 1; i++) { var diff = Math.abs(arr[i] - arr[i + 1]); if (diff > max_diff) max_diff = diff; } // Return the maximum absolute difference return max_diff;}// Driver Codevar arr = [ -1, -1, 11, -1, 3, -1 ];var n = arr.length;// Function calldocument.write(maximumAbsolute(arr, n));// This code is contributed by umadevi9616 </script> |
4
Time complexity: O(N), The time complexity of the given program is O(n), where n is the size of the input array. This is because the program iterates through the input array twice in two separate for-loops, which have a time complexity of O(n) each.
Auxiliary Space: O(1), The space complexity of the given program is O(1), which means it uses a constant amount of memory regardless of the size of the input array. This is because the program does not create any new data structures that depend on the size of the input array. It only uses a fixed number of integer variables to store the minimum, maximum, and common values, as well as the maximum absolute difference.
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