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> |
0 1 2 3 4
Time Complexity: O(N)
Auxiliary Space: O(N), since N extra space has been taken.