Given an array of N elements, the task is to minimize the maximum difference of adjacent elements by inserting at most K elements in the array.
Examples:
Input: arr = [2, 6, 8] K = 1
Output: 2
Explanation:
After insertion of 4 in between 2 and 6, the array becomes [2, 4, 6, 8]. In this case the maximum difference between any adjacent element is 2, which is the minimum that can be.
Input: arr = [3, 12] K = 2
Output: 3
Explanation:
After insertion of 6 and 9 in between 3 and 12, the array becomes [3, 6, 9, 12]. In this case the maximum difference between any adjacent element is 3, which is the minimum that can be.
Approach: In order to solve this problem, we are using the following Binary Search based approach:
- Find the maximum difference between any two adjacent element in the array and store it in a variable, say worst.
- Search from best(1 initially) to worst and for every mid value find the number of insertions required.
- Whenever the number of insertions is greater than K for a particular value of mid, search between [mid + 1, worst], that is the higher half. Otherwise search between [best, mid-1], that is the lower half to check if the maximum difference can be further minimized with at most K insertions.
- The final worst value after termination of the loop gives the answer.
Below code is the implementation of the above approach:
C++
// C++ Program to find the minimum of maximum// differerence between adjacent elements// after at most K insertions#include <bits/stdc++.h>using namespace std;int minMaxDiff(int arr[], int n, int k){ int max_adj_dif = INT_MIN; // Calculate the maximum // adjacent difference for (int i = 0; i < n - 1; i++) max_adj_dif = max(max_adj_dif, abs(arr[i] - arr[i + 1])); // If the maximum adjacent // difference is already zero if (max_adj_dif == 0) return 0; // best and worst specifies // range of the maximum // adjacent difference int best = 1; int worst = max_adj_dif; int mid, required; while (best < worst) { mid = (best + worst) / 2; // To store the no of insertions // required for respective // values of mid required = 0; for (int i = 0; i < n - 1; i++) { required += (abs(arr[i] - arr[i + 1]) - 1) / mid; } // If the number of insertions // required exceeds K if (required > k) best = mid + 1; // Otherwise else worst = mid; } return worst;}// Driver codeint main(){ int arr[] = { 3, 12, 25, 50 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 7; cout << minMaxDiff(arr, n, k); return 0;} |
Java
// Java program to find the minimum // of maximum difference between // adjacent elements after at most// K insertionsimport java.util.*;class GFG{ static int minMaxDiff(int arr[], int n, int k){ int max_adj_dif = Integer.MIN_VALUE; // Calculate the maximum // adjacent difference for(int i = 0; i < n - 1; i++) max_adj_dif = Math.max(max_adj_dif, Math.abs(arr[i] - arr[i + 1])); // If the maximum adjacent // difference is already zero if (max_adj_dif == 0) return 0; // best and worst specifies // range of the maximum // adjacent difference int best = 1; int worst = max_adj_dif; int mid, required; while (best < worst) { mid = (best + worst) / 2; // To store the no of insertions // required for respective // values of mid required = 0; for(int i = 0; i < n - 1; i++) { required += (Math.abs(arr[i] - arr[i + 1]) - 1) / mid; } // If the number of insertions // required exceeds K if (required > k) best = mid + 1; // Otherwise else worst = mid; } return worst;}// Driver codepublic static void main(String[] args){ int arr[] = { 3, 12, 25, 50 }; int n = arr.length; int k = 7; System.out.println(minMaxDiff(arr, n, k));}}// This code is contributed by ANKITKUMAR34 |
Python 3
# Python3 program to find the minimum # of maximum difference between # adjacent elements after at most# K insertionsdef minMaxDiff(arr, n, k): max_adj_dif = float('-inf'); # Calculate the maximum # adjacent difference for i in range(n - 1): max_adj_dif = max(max_adj_dif, abs(arr[i] - arr[i + 1])); # If the maximum adjacent # difference is already zero if (max_adj_dif == 0): return 0; # best and worst specifies # range of the maximum # adjacent difference best = 1; worst = max_adj_dif; while (best < worst): mid = (best + worst) // 2; # To store the no of insertions # required for respective # values of mid required = 0 for i in range(n - 1): required += (abs(arr[i] - arr[i + 1]) - 1) // mid # If the number of insertions # required exceeds K if (required > k): best = mid + 1; # Otherwise else: worst = mid return worst# Driver codearr = [ 3, 12, 25, 50 ]n = len(arr)k = 7print(minMaxDiff(arr, n, k))# This code is contributed by ANKITKUMAR34 |
C#
// C# program to find the minimum // of maximum difference between // adjacent elements after at most// K insertionsusing System;class GFG{ static int minMaxDiff(int []arr, int n, int k){ int max_adj_dif = int.MinValue; // Calculate the maximum // adjacent difference for(int i = 0; i < n - 1; i++) max_adj_dif = Math.Max(max_adj_dif, Math.Abs(arr[i] - arr[i + 1])); // If the maximum adjacent // difference is already zero if (max_adj_dif == 0) return 0; // best and worst specifies // range of the maximum // adjacent difference int best = 1; int worst = max_adj_dif; int mid, required; while (best < worst) { mid = (best + worst) / 2; // To store the no of insertions // required for respective // values of mid required = 0; for(int i = 0; i < n - 1; i++) { required += (Math.Abs(arr[i] - arr[i + 1]) - 1) / mid; } // If the number of insertions // required exceeds K if (required > k) best = mid + 1; // Otherwise else worst = mid; } return worst;}// Driver codepublic static void Main(String[] args){ int []arr = { 3, 12, 25, 50 }; int n = arr.Length; int k = 7; Console.WriteLine(minMaxDiff(arr, n, k));}}// This code is contributed by Princi Singh |
Javascript
<script>// javascript Program to find the minimum of maximum// differerence between adjacent elements// after at most K insertionsfunction minMaxDiff(arr, n, k){ var max_adj_dif = -1000000000; // Calculate the maximum // adjacent difference for (var i = 0; i < n - 1; i++) max_adj_dif = Math.max(max_adj_dif, Math.abs(arr[i] - arr[i + 1])); // If the maximum adjacent // difference is already zero if (max_adj_dif == 0) return 0; // best and worst specifies // range of the maximum // adjacent difference var best = 1; var worst = max_adj_dif; var mid, required; while (best < worst) { mid = (best + worst) / 2; // To store the no of insertions // required for respective // values of mid required = 0; for (var i = 0; i < n - 1; i++) { required += parseInt((Math.abs(arr[i] - arr[i + 1]) - 1) / mid); } // If the number of insertions // required exceeds K if (required > k) best = mid + 1; // Otherwise else worst = mid; } return worst;}// Driver codevar arr = [ 3, 12, 25, 50 ];var n = arr.length;var k = 7;document.write( minMaxDiff(arr, n, k));</script> |
Output:
5
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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