Given an array arr[] of size N, the task is to print the distance of every array element from its next greater element. For array elements having no next greater element, print 0.
Examples:
Input: arr[] = {73, 74, 75, 71, 69, 72, 76, 73}
Output: {1, 1, 4, 2, 1, 1, 0, 0}
Explanation:
The next greater element for 73 is 74, which is at position 1. Distance = 1 – 0 = 1
The next greater element for 74 is 75, which is at position 2. Distance = 2 – 1 = 1
The next greater element for 75 is 76, which is at position 6. Distance = 6 – 2 = 4
The next greater element for 71 is 72, which is at position 5. Distance = 5 – 3 = 2
The next greater element for 69 is 72, which is at position 5. Distance = 5 – 4 = 1
The next greater element for 72 is 76, which is at position 6. Distance = 6 – 5 = 1
No, next greater element for 76. Distance = 0
No, next greater element for 73. Distance = 0Input: arr[] = {5, 4, 3, 2, 1}
Output: {0, 0, 0, 0, 0}
Naive Approach: The simplest approach is to traverse the array and for every array element, traverse to its right to obtain its next greater element and calculate the difference between the indices.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Stack to find the next greater element.
Below are the steps:
- Maintain a Stack which will contain the elements in non-increasing order.
- Check if the current element arr[i]is greater than the element at the top of the stack.
- Keep popping all the elements from the stack one by one from the top, that are found to be smaller than arr[i] and calculate the distance for each of them as the difference of current index and the index of the popped element.
- Push the current element into the stack and repeat the above steps.
Below is the implementation of the above approach:
C++
// C++ implementation of the // above approach #include<bits/stdc++.h>using namespace std;vector<int> mindistance(vector<int> arr){ int N = arr.size(); // Stores the required distances vector<int> ans(N); int st = 0; // Maintain a stack of elements // in non-increasing order for(int i = 0; i < N - 1; i++) { if (arr[i] < arr[i + 1]) { ans[i] = 1; } else { st = i + 1; while (st <= N - 1) { if (arr[i] < arr[st]) { ans[i] = st - i; break; } else { st++; } } } } return ans;}// Driver codeint main(){ vector<int> arr = { 73, 74, 75, 71, 69, 72, 76, 73 }; vector<int> x = mindistance(arr); cout << "["; for(int i = 0; i < x.size(); i++) { if (i == x.size() - 1) cout << x[i]; else cout << x[i] << ", "; } cout << "]";}// This code is contributed by SURENDRA_GANGWAR |
Java
// Java implementation of the // above approach import java.io.*;class GFG{ public static int[] mindistance(int[] arr){ int N = arr.length; // Stores the required distances int[] ans = new int[N]; int st = 0; // Maintain a stack of elements // in non-increasing order for(int i = 0; i < N - 1; i++) { if (arr[i] < arr[i + 1]) { ans[i] = 1; } else { st = i + 1; while (st <= N - 1) { if (arr[i] < arr[st]) { ans[i] = st - i; break; } else { st++; } } } } return ans;}// Driver codepublic static void main(String[] args){ int arr[] = new int[]{ 73, 74, 75, 71, 69, 72, 76, 73 }; int x[] = mindistance(arr); System.out.print("["); for(int i = 0; i < x.length; i++) System.out.print(x[i]+", "); System.out.print("]");}}// This code is contributed by sai-sampath mahajan// and ramprasad kondoju |
Python3
# Python3 implementation of the# above approachdef mindistance(arr, N): if N <= 1: return [0] # Stores the required distances ans = [0 for i in range(N)] st = [0] # Maintain a stack of elements # in non-increasing order for i in range(1, N): # If the current element exceeds # the element at the top of the stack while(st and arr[i] > arr[st[-1]]): pos = st.pop() ans[pos] = i - pos # Push the current index to the stack st.append(i) return ans# Given arrayarr = [73, 74, 75, 71, 69, 72, 76, 73]N = len(arr)# Function callprint(mindistance(arr, N)) |
C#
// C# implementation of the // above approach using System;class GFG{ public static int[] mindistance(int[] arr){ int N = arr.Length; // Stores the required distances int[] ans = new int[N]; int st = 0; // Maintain a stack of elements // in non-increasing order for(int i = 0; i < N - 1; i++) { if (arr[i] < arr[i + 1]) { ans[i] = 1; } else { st = i + 1; while (st <= N - 1) { if (arr[i] < arr[st]) { ans[i] = st - i; break; } else { st++; } } } } return ans;}// Driver codepublic static void Main(String[] args){ int []arr = new int[]{ 73, 74, 75, 71, 69, 72, 76, 73 }; int []x = mindistance(arr); Console.Write("["); for(int i = 0; i < x.Length; i++) Console.Write(x[i]+", "); Console.Write("]");}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program to implement// the above approachfunction mindistance(arr){ let N = arr.length; // Stores the required distances let ans = []; let st = 0; // Maintain a stack of elements // in non-increasing order for(let i = 0; i < N - 1; i++) { if (arr[i] < arr[i + 1]) { ans[i] = 1; } else { st = i + 1; while (st <= N - 1) { if (arr[i] < arr[st]) { ans[i] = st - i; break; } else { st++; } } } } return ans;}// Driver code let arr = [ 73, 74, 75, 71, 69, 72, 76, 73 ]; let x = mindistance(arr); document.write("["); for(let i = 0; i < x.length; i++) document.write(x[i]+", "); document.write("]");// This code is contributed by target_2.</script> |
[1, 1, 4, 2, 1, 1, 0, 0]
Time Complexity: O(N) //since one traversal of the array is required to complete all operations hence overall time required by the algorithm is linear
Auxiliary Space: O(N)// an extra array is used and in the worst case all elements will be stored inside it the hence algorithm takes up linear space
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