Given an array, arr[] of size N, the task is to find the number of sub-arrays ending with arr[i] and arr[i] is the minimum element of that sub-array.
Examples:
Input: arr[] = {3, 1, 2, 4}
Output: 1 2 1 1
Explanation:
Subarrays ending with 3 where 3 is the minimum element = {3}
Subarrays ending with 1 where 1 is the minimum element = {3, 1}, {1}
Subarrays ending with 2 where 2 is the minimum element = {2}
Subarrays ending with 4 where 4 is the minimum element = {4}
Input: arr[] = {5, 4, 3, 2, 1}
Output: 1 2 3 4 5
Approach: The idea is to use the approach used to find Next Greater Element by maintaining a stack. Step-wise approach for the problem is:
- Push the first element (arr[0]) of the array with count as 1 in the stack because first element will be the sub-array itself ending with current element arr[0] and minimum of the subarray
- Then for each element arr[i] in the array-
- Pop the elements from the stack until the top of the stack is greater than the current element and add the count of the popped element in the count of current element.
- Push current element and the count as a pair in the stack.
For example: For arr[] = {3, 1, 2, 4},
Below is the implementation of the above approach:
C++
// C++ implementation to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarray#include <bits/stdc++.h>using namespace std;// Function to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarrayint min_subarray(int a[], int n){ stack<pair<int, int> > st; for (int i = 0; i < n; ++i) { // There exists a subarray of // size 1 for each element int count = 1; // Remove all greater elements while (!st.empty() && st.top().first > a[i]) { // Increment the count count += st.top().second; // Remove the element st.pop(); } // Push the current element // and it's count st.push({ a[i], count }); cout << count << " "; }}// Driver Codeint main(){ int a[] = {5, 4, 3, 2, 1}; int n = sizeof(a) / sizeof(a[0]); min_subarray(a, n); return 0;} |
Java
// Java implementation to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarray import java.util.*;import java.lang.*;import java.io.*;class Main{ static class Pair { int first; int second; public Pair(int x, int y) { this.first = x; this.second = y; } } // Function to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarraystatic void min_subarray(int []a, int n){ Stack<Pair> st = new Stack<Pair>(); for (int i = 0; i < n; ++i) { // There exists a subarray of // size 1 for each element int count = 1; // Remove all greater elements while (st.empty() == false && st.peek().first > a[i]) { // Increment the count count += st.peek().second; // Remove the element st.pop(); } // Push the current element // and it's count st.push(new Pair (a[i], count )); System.out.print(count + " "); }}// Driver Codepublic static void main(String []args){ int []a = {5, 4, 3, 2, 1}; int n = a.length; min_subarray(a, n);}}// This code is contributed by tufan_gupta2000 |
Python3
# Python3 implementation to find the number # of sub-arrays ending with arr[i] which # is the minimum element of the subarray # Function to find the number # of sub-arrays ending with arr[i] which # is the minimum element of the subarray def min_subarray(a, n) : st = []; for i in range(n) : # There exists a subarray of # size 1 for each element count = 1; # Remove all greater elements while len(st) != 0 and st[-1][0] > a[i] : # Increment the count count += st[-1][1]; # Remove the element st.pop(); # Push the current element # and it's count st.append(( a[i], count )); print(count,end= " "); # Driver Code if __name__ == "__main__" : a = [5, 4, 3, 2, 1]; n = len(a); min_subarray(a, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarray using System;using System.Collections.Generic;class GFG{ class Pair { public int first; public int second; public Pair(int x, int y) { this.first = x; this.second = y; } } // Function to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarraystatic void min_subarray(int []a, int n){ Stack<Pair> st = new Stack<Pair>(); for (int i = 0; i < n; ++i) { // There exists a subarray of // size 1 for each element int count = 1; // Remove all greater elements while (st.Count != 0 && st.Peek().first > a[i]) { // Increment the count count += st.Peek().second; // Remove the element st.Pop(); } // Push the current element // and it's count st.Push(new Pair (a[i], count )); Console.Write(count + " "); }}// Driver Codepublic static void Main(String []args){ int []a = {5, 4, 3, 2, 1}; int n = a.Length; min_subarray(a, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarray// Function to find the number// of sub-arrays ending with arr[i] which// is the minimum element of the subarrayfunction min_subarray(a, n){ var st = []; for (var i = 0; i < n; ++i) { // There exists a subarray of // size 1 for each element var count = 1; // Remove all greater elements while (st.length!=0 && st[st.length-1][0] > a[i]) { // Increment the count count += st[st.length-1][1]; // Remove the element st.pop(); } // Push the current element // and it's count st.push([a[i], count]); document.write( count + " "); }}// Driver Codevar a = [5, 4, 3, 2, 1];var n = a.length;min_subarray(a, n);// This code is contributed by itsok.</script> |
Output:
1 2 3 4 5
Time Complexity : O(n) ,where n is size of given array.
Space Complexity : O(n)
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