Given an array arr[] of size N and an integer K, the task is to find all the indices in the given array having at least K non-increasing elements before and K non-decreasing elements after them.
Examples:
Input: arr[] = {1, 1, 1, 1, 1}, K = 0
Output: 0 1 2 3 4
Explanation: Since K equals 0, every index satisfies the condition.Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 2
Output: -1
Explanation: No index has 2 non-increasing before it and 2 non-decreasing elements after it.
Approach: The solution can be found by using the concept of prefix and suffix array. Follow the steps mentioned below:
- Form the prefix[] array where prefix[i] represents the number of elements before i which obeys non-increasing order.
- Form the suffix[] array where suffix[i] represents the number of elements after i which obeys non-decreasing order.
- Now only those indexes should be included in the answer for which, both prefix[i] and suffix[i] are greater than or equal to K.
Below is the implementation of the above approach.
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to find all the indicesvector<int> findIndices(int arr[], int K, int N){ vector<int> prefix(N), suffix(N); vector<int> ans; prefix[0] = 1; for (int i = 1; i < N; i++) { if (arr[i] <= arr[i - 1]) prefix[i] = prefix[i - 1] + 1; else prefix[i] = 1; } suffix[N - 1] = 1; for (int i = N - 2; i >= 0; i--) { if (arr[i] <= arr[i + 1]) suffix[i] = suffix[i + 1] + 1; else suffix[i] = 1; } for (int i = 0; i < N; i++) { if (K == 0 || i - 1 >= 0 && i + 1 < N && prefix[i - 1] >= K && suffix[i + 1] >= K) ans.push_back(i); } return ans;}// Driver codeint main(){ int arr[] = { 1, 1, 1, 1, 1 }; int K = 0; int N = sizeof(arr) / sizeof(arr[0]); vector<int> ans = findIndices(arr, K, N); for (int i = 0; i < ans.size(); i++) { cout << ans[i] << " "; } if (ans.size() == 0) cout << "-1"; return 0;} |
Java
// Java code for the above approachimport java.util.ArrayList;class GFG { // Function to find all the indices static ArrayList<Integer> findIndices(int[] arr, int K, int N) { int[] prefix = new int[N]; int[] suffix = new int[N]; ArrayList<Integer> ans = new ArrayList<Integer>(); prefix[0] = 1; for (int i = 1; i < N; i++) { if (arr[i] <= arr[i - 1]) prefix[i] = prefix[i - 1] + 1; else prefix[i] = 1; } suffix[N - 1] = 1; for (int i = N - 2; i >= 0; i--) { if (arr[i] <= arr[i + 1]) suffix[i] = suffix[i + 1] + 1; else suffix[i] = 1; } for (int i = 0; i < N; i++) { if (K == 0 || i - 1 >= 0 && i + 1 < N && prefix[i - 1] >= K && suffix[i + 1] >= K) ans.add(i); } return ans; } // Driver code public static void main(String args[]) { int[] arr = { 1, 1, 1, 1, 1 }; int K = 0; int N = arr.length; ArrayList<Integer> ans = findIndices(arr, K, N); for (int i = 0; i < ans.size(); i++) { System.out.print(ans.get(i) + " "); } if (ans.size() == 0) System.out.println("-1"); }}// This code is contributed by gfgking |
Python3
# Python code for the above approach# Function to find all the indicesdef findIndices(arr, K, N): prefix = [0] * N suffix = [0] * N ans = [] prefix[0] = 1 for i in range(1, N): if (arr[i] <= arr[i - 1]): prefix[i] = prefix[i - 1] + 1 else: prefix[i] = 1 suffix[N - 1] = 1 for i in range(N - 2, 1, -1): if (arr[i] <= arr[i + 1]): suffix[i] = suffix[i + 1] + 1 else: suffix[i] = 1 for i in range(N): if (K == 0 or i - 1 >= 0 and i + 1 < N and prefix[i - 1] >= K and suffix[i + 1] >= K): ans.append(i) return ans# Driver codearr = [1, 1, 1, 1, 1]K = 0N = len(arr)ans = findIndices(arr, K, N)for i in range(len(ans)): print(ans[i], end=" ")if (len(ans) == 0): print("-1")# This code is contributed by Saurabh Jaiswal |
C#
// C# code for the above approachusing System;using System.Collections.Generic;class GFG { // Function to find all the indices static List<int> findIndices(int[] arr, int K, int N) { int[] prefix = new int[N]; int[] suffix = new int[N]; List<int> ans = new List<int>(); prefix[0] = 1; for (int i = 1; i < N; i++) { if (arr[i] <= arr[i - 1]) prefix[i] = prefix[i - 1] + 1; else prefix[i] = 1; } suffix[N - 1] = 1; for (int i = N - 2; i >= 0; i--) { if (arr[i] <= arr[i + 1]) suffix[i] = suffix[i + 1] + 1; else suffix[i] = 1; } for (int i = 0; i < N; i++) { if (K == 0 || i - 1 >= 0 && i + 1 < N && prefix[i - 1] >= K && suffix[i + 1] >= K) ans.Add(i); } return ans; } // Driver code public static void Main() { int[] arr = { 1, 1, 1, 1, 1 }; int K = 0; int N = arr.Length; List<int> ans = findIndices(arr, K, N); for (int i = 0; i < ans.Count; i++) { Console.Write(ans[i] + " "); } if (ans.Count == 0) Console.Write("-1"); }}// This code is contributed by ukasp |
Javascript
<script> // JavaScript code for the above approach // Function to find all the indices const findIndices = (arr, K, N) => { let prefix = new Array(N).fill(0); let suffix = new Array(N).fill(0); let ans = []; prefix[0] = 1; for (let i = 1; i < N; i++) { if (arr[i] <= arr[i - 1]) prefix[i] = prefix[i - 1] + 1; else prefix[i] = 1; } suffix[N - 1] = 1; for (let i = N - 2; i >= 0; i--) { if (arr[i] <= arr[i + 1]) suffix[i] = suffix[i + 1] + 1; else suffix[i] = 1; } for (let i = 0; i < N; i++) { if (K == 0 || i - 1 >= 0 && i + 1 < N && prefix[i - 1] >= K && suffix[i + 1] >= K) ans.push(i); } return ans; } // Driver code let arr = [1, 1, 1, 1, 1]; let K = 0; let N = arr.length; let ans = findIndices(arr, K, N); for (let i = 0; i < ans.length; i++) { document.write(`${ans[i]} `); } if (ans.length == 0) cout << "-1";// This code is contributed by rakeshsahni</script> |
Output
0 1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(N), since N extra space has been taken.
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