Given an array arr[] consisting of integers, the task is to split the given array into two sub-arrays such that the difference between their maximum elements is minimum.
Example:
Input: arr[] = {7, 9, 5, 10}
Output: 1
Explanation:
The subarrays are {5, 10} and {7, 9} with the difference between their maximums = 10 – 9 = 1.
Input: arr[] = {6, 6, 6}
Output: 0
Approach:
We can observe that we need to split the array into two subarrays such that:
- If the maximum element occurs more than once in the array, it needs to be present in both the subarrays at least once.
- Otherwise, the largest and the second-largest elements should be present in different subarrays.
This ensures that the difference between the maximum elements of the two subarrays is maximized.
Hence, we need to sort the array, and then the difference between the largest 2 elements, i.e. arr[n – 1] and arr[n – 2], is the required answer.
Below is the implementation of the above approach:
C++
// C++ Program to split a given// array such that the difference// between their maximums is minimized.#include <bits/stdc++.h>using namespace std;int findMinDif(int arr[], int N){ // Sort the array sort(arr, arr + N); // Return the difference // between two highest // elements return (arr[N - 1] - arr[N - 2]);}// Driver Programint main(){ int arr[] = { 7, 9, 5, 10 }; int N = sizeof(arr) / sizeof(arr[0]); cout << findMinDif(arr, N); return 0;} |
Java
// Java Program to split a given array // such that the difference between // their maximums is minimized.import java.util.*;class GFG{static int findMinDif(int arr[], int N){ // Sort the array Arrays.sort(arr); // Return the difference between // two highest elements return (arr[N - 1] - arr[N - 2]);}// Driver codepublic static void main(String[] args){ int arr[] = { 7, 9, 5, 10 }; int N = arr.length; System.out.println(findMinDif(arr, N));}}// This code is contributed by offbeat |
Python3
# Python3 Program to split a given # array such that the difference # between their maximums is minimized. def findMinDif(arr, N): # Sort the array arr.sort() # Return the difference # between two highest # elements return (arr[N - 1] - arr[N - 2])# Driver Program arr = [ 7, 9, 5, 10 ]N = len(arr)print(findMinDif(arr, N))# This code is contributed by yatinagg |
C#
// C# Program to split a given array // such that the difference between // their maximums is minimized.using System;class GFG{static int findMinDif(int []arr, int N){ // Sort the array Array.Sort(arr); // Return the difference between // two highest elements return (arr[N - 1] - arr[N - 2]);}// Driver codepublic static void Main(){ int []arr = { 7, 9, 5, 10 }; int N = arr.Length; Console.Write(findMinDif(arr, N));}}// This code is contributed by Code_Mech |
Javascript
<script>// javascript Program to split a given array // such that the difference between // their maximums is minimized. function findMinDif(arr , N) { // Sort the array arr.sort((a,b)=>a-b); // Return the difference between // two highest elements return (arr[N - 1] - arr[N - 2]); } // Driver code var arr = [ 7, 9, 5, 10 ]; var N = arr.length; document.write(findMinDif(arr, N));// This code contributed by gauravrajput1</script> |
Output:
1
Time complexity: O(N*log(N)), N is the number of elements of the array.
Auxiliary Space: O(1)
Another Approach: We can optimize the above code by removing the sort function used above. As the answer is basically the difference between the two greatest elements of the array, so we can traverse the array and can find two greatest elements in O(n) time.
Below is the code for the given approach:
C++
// C++ Program to split a given// array such that the difference// between their maximums is minimized.#include <bits/stdc++.h>using namespace std;int findMinDif(int arr[], int n){ int first_max = INT_MIN; int second_max = INT_MIN; for (int i = 0; i < n ; i ++) { // If current element is greater than first // then update both first and second if (arr[i] > first_max) { second_max = first_max; first_max = arr[i]; } // If arr[i] is less and equal to first_max // but greater than second_max // then update the second_max else if (arr[i] > second_max) second_max = arr[i]; } // Return the difference // between two highest // elements return first_max-second_max; }// Driver codeint main(){ int arr[] = { 7, 9, 5, 10 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findMinDif(arr, n) << endl; return 0;}// This code is contributed by Pushpesh Raj |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.*;class GFG { public static int findMinDif(int[] arr, int n) { int first_max = Integer.MIN_VALUE; int second_max = Integer.MIN_VALUE; for (int i = 0; i < n; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first_max) { second_max = first_max; first_max = arr[i]; } // If arr[i] is less and equal to first_max // but greater than second_max // then update the second_max else if (arr[i] > second_max) second_max = arr[i]; } // Return the difference // between two highest // elements return first_max - second_max; } public static void main(String[] args) { int[] arr = { 7, 9, 5, 10 }; int n = arr.length; System.out.println(findMinDif(arr, n)); }}// This code is contributed by akashish__ |
Python3
def findMinDif(arr, n): first_max = -2147483647 second_max = -2147483647 #for (int i = 0; i < n ; i ++) for i in range(0,n): # If current element is greater than first # then update both first and second if (arr[i] > first_max): second_max = first_max first_max = arr[i] # If arr[i] is less and equal to first_max # but greater than second_max # then update the second_max elif (arr[i] > second_max): second_max = arr[i] # Return the difference # between two highest # elements return first_max-second_max# Driver codearr = [7, 9, 5, 10 ]n = len(arr)print(findMinDif(arr, n))# This code is contributed by akashish__ |
C#
using System;public class GFG { public static int findMinDif(int[] arr, int n) { int first_max = Int32.MinValue; int second_max = Int32.MinValue; for (int i = 0; i < n; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first_max) { second_max = first_max; first_max = arr[i]; } // If arr[i] is less and equal to first_max // but greater than second_max // then update the second_max else if (arr[i] > second_max) second_max = arr[i]; } // Return the difference // between two highest // elements return first_max - second_max; } static public void Main() { int[] arr = { 7, 9, 5, 10 }; int n = arr.Length; Console.WriteLine(findMinDif(arr, n)); }}// This code is contributed by akashish__ |
Javascript
<script>// Javascript Program to split a given// array such that the difference// between their maximums is minimized.function findMinDif(arr,n){ let first_max = Number.MIN_VALUE; let second_max = Number.MIN_VALUE; for (let i = 0; i < n ; i ++) { // If current element is greater than first // then update both first and second if (arr[i] > first_max) { second_max = first_max; first_max = arr[i]; } // If arr[i] is less and equal to first_max // but greater than second_max // then update the second_max else if (arr[i] > second_max) second_max = arr[i]; } // Return the difference // between two highest // elements return first_max-second_max; }// Driver codelet arr =[ 7, 9, 5, 10 ];let n = arr.length;console.log(findMinDif(arr, n));// This code is contributed by akashish__</script> |
Output
1
Time Complexity: O(n)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
