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++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find all the indices
vector<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 code
int 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 code for the above approach
import 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
# Python code for the above approach
# Function to find all the indices
def 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 code
arr = [1, 1, 1, 1, 1]
K = 0
N = 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# code for the above approach
using 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
<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.
