Given an array arr[] consisting of N distinct integers, the task is to find the number of subarrays which contains both the maximum and the minimum element from the given array.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 1
Explanation:
Only a single subarray {1, 2, 3, 4} consists of both the maximum (= 4) and the minimum (= 1) array elements.Input: arr[] = {4, 1, 2, 3}
Output: 3
Explanation:
Subarrays {4, 1} , {4, 1, 2}, {4, 1, 2, 3} consists of both the maximum(= 4) and the minimum(= 1) array elements .
Naive Approach: The simplest approach is to first, traverse the array and find the maximum and minimum of the array and then generate all possible subarrays of the given array. For each subarray, check if it contains both the maximum and the minimum array element. For all such subarrays, increase the count by 1. Finally, print the count of such subarrays.
Code-
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count subarray// containing both maximum and// minimum array elementsint countSubArray(int arr[], int n){ int maxi=INT_MIN; int mini=INT_MAX; for(int i=0;i<n;i++){ maxi=max(maxi,arr[i]); mini=min(mini,arr[i]); } int count=0; for(int i=0;i<n;i++){ int temp_max=arr[i]; int temp_min=arr[i]; for(int j=i;j<n;j++){ if(arr[j]>temp_max){temp_max=arr[j];} if(arr[j]<temp_min){temp_min=arr[j];} //Checking that our subarray will contain //maximum and minimum element of array if((mini==temp_min) && (maxi==temp_max) ){ count++; } } } return count;}// Driver Codeint main(){ int arr[] = { 4,1,2,3 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << countSubArray(arr, n); return 0;} |
Javascript
// Function to count subarray// containing both maximum and// minimum array elementsfunction countSubArray(arr) { let maxi = Number.MIN_SAFE_INTEGER; let mini = Number.MAX_SAFE_INTEGER; for (let i = 0; i < arr.length; i++) { maxi = Math.max(maxi, arr[i]); mini = Math.min(mini, arr[i]); } let count = 0; for (let i = 0; i < arr.length; i++) { let temp_max = arr[i]; let temp_min = arr[i]; for (let j = i; j < arr.length; j++) { if (arr[j] > temp_max) { temp_max = arr[j]; } if (arr[j] < temp_min) { temp_min = arr[j]; } // Checking that our subarray will contain // maximum and minimum element of array if (mini == temp_min && maxi == temp_max) { count++; } } } return count;}// Driver Codelet arr = [4, 1, 2, 3];// Function callconsole.log(countSubArray(arr)); |
Output-
3
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: Follow the steps below to optimize the above approach:
- Find the index of the maximum and minimum elements. Let i and j be the respective indices such that i < j.
- All the subarray which starts from indices up to i and ends at indices after j will contain the maximum as well as the minimum array element.
- Therefore, the possible indices for the starting index of the subarray are [0, i] (total = i + 1 ).
- Therefore, the possible indices for the ending index of the subarray are [j, N – 1] (total = N – j).
- Therefore, the count of subarrays is given by (i + 1) * ( N – j).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count subarray// containing both maximum and// minimum array elementsint countSubArray(int arr[], int n){ // If the length of the // array is less than 2 if (n < 2) return n; // Find the index of maximum element int i = max_element(arr, arr + n) - arr; // Find the index of minimum element int j = min_element(arr, arr + n) - arr; // If i > j, then swap // the value of i and j if (i > j) swap(i, j); // Return the answer return (i + 1) * (n - j);}// Driver Codeint main(){ int arr[] = { 4, 1, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << countSubArray(arr, n); return 0;} |
Java
// Java program for the above approachimport java.util.*;import java.lang.*;class GFG{ // Function to count subarray // containing both maximum and // minimum array elements static int countSubArray(int arr[], int n) { // If the length of the // array is less than 2 if (n < 2) return n; // Find the index of maximum element int i = max_element(arr); // Find the index of minimum element int j = min_element(arr); // If i > j, then swap // the value of i and j if (i > j) { int tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; } // Return the answer return (i + 1) * (n - j); } // Function to return max_element indexstatic int max_element(int[] arr){ int idx = 0; int max = arr[0]; for(int i = 1; i < arr.length; i++) { if(max < arr[i]) { max = arr[i]; idx = i; } } return idx; }// Function to return min_element indexstatic int min_element(int[] arr){ int idx = 0; int min = arr[0]; for(int i = 1; i < arr.length; i++) { if (arr[i] < min) { min = arr[i]; idx = i; } } return idx; }// Driver Codepublic static void main (String[] args){ int arr[] = { 4, 1, 2, 3 }; int n = arr.length; // Function call System.out.println(countSubArray(arr, n)); }}// This code is contributed by offbeat |
Python3
# Python3 program for # the above approach# Function to count subarray# containing both maximum and# minimum array elementsdef countSubArray(arr, n): # If the length of the # array is less than 2 if (n < 2): return n; # Find the index of # maximum element i = max_element(arr); # Find the index of # minimum element j = min_element(arr); # If i > j, then swap # the value of i and j if (i > j): tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; # Return the answer return (i + 1) * (n - j);# Function to return # max_element indexdef max_element(arr): idx = 0; max = arr[0]; for i in range(1, len(arr)): if (max < arr[i]): max = arr[i]; idx = i; return idx;# Function to return # min_element indexdef min_element(arr): idx = 0; min = arr[0]; for i in range(1, len(arr)): if (arr[i] < min): min = arr[i]; idx = i; return idx;# Driver Codeif __name__ == '__main__': arr = [4, 1, 2, 3]; n = len(arr); # Function call print(countSubArray(arr, n));# This code is contributed by Rajput-Ji |
C#
// C# program for // the above approachusing System;class GFG{ // Function to count subarray // containing both maximum and // minimum array elements static int countSubArray(int []arr, int n) { // If the length of the // array is less than 2 if (n < 2) return n; // Find the index of maximum element int i = max_element(arr); // Find the index of minimum element int j = min_element(arr); // If i > j, then swap // the value of i and j if (i > j) { int tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; } // Return the answer return (i + 1) * (n - j); } // Function to return max_element indexstatic int max_element(int[] arr){ int idx = 0; int max = arr[0]; for(int i = 1; i < arr.Length; i++) { if(max < arr[i]) { max = arr[i]; idx = i; } } return idx; }// Function to return min_element indexstatic int min_element(int[] arr){ int idx = 0; int min = arr[0]; for(int i = 1; i < arr.Length; i++) { if (arr[i] < min) { min = arr[i]; idx = i; } } return idx; }// Driver Codepublic static void Main(String[] args){ int []arr = {4, 1, 2, 3}; int n = arr.Length; // Function call Console.WriteLine(countSubArray(arr, n)); }}// This code is contributed by shikhasingrajput |
Javascript
<script>// javascript program for the// above approach// Function to count subarray// containing both maximum and// minimum array elementsfunction countSubArray(arr, n){ // If the length of the // array is less than 2 if (n < 2) return n; // Find the index of maximum element let i = max_element(arr); // Find the index of minimum element let j = min_element(arr); // If i > j, then swap // the value of i and j if (i > j) { let tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; } // Return the answer return (i + 1) * (n - j);} // Function to return max_element indexfunction max_element(arr){ let idx = 0; let max = arr[0]; for(let i = 1; i < arr.length; i++) { if(max < arr[i]) { max = arr[i]; idx = i; } } return idx;} // Function to return min_element indexfunction min_element(arr){ let idx = 0; let min = arr[0]; for(let i = 1; i < arr.length; i++) { if (arr[i] < min) { min = arr[i]; idx = i; } } return idx;} // Driver Code let arr = [ 4, 1, 2, 3 ]; let n = arr.length; // Function call document.write(countSubArray(arr, n)); // This code is contributed by avijitmondal1998</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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