Subarray permutation that satisfies the given condition
Last Updated :
12 Jul, 2025
Given a permutation of integers from 1 to N and an integer M, the task is to check if any subarray of the given permutation is a permutation of the integers from 1 to M.
Examples:
Input: arr[] = {4, 5, 1, 3, 2, 6}, M = 3
Output: Yes
{4, 5, 1, 3, 2, 6} is the required subarray.
Input: arr[] = {4, 5, 1, 3, 2, 6}, M = 4
Output: No
Naive approach: A naive approach would be to generate all the M-sized subarrays and see if any such subarray exists. But if the given permutation is too large, this approach will be time-consuming as it runs in O(N3).
Efficient approach: A better solution is to use Hashing.
- From the main permutation, the positions of each integer are stored in a map/dictionary.
- Now, observe that if there exists a subarray which is a permutation from 1 to m, then all numbers in range [1, m] will occupy m consecutive places in the main permutation, either in a sorted or random manner.
- Their positions also should come as m-consecutive numbers, when sorted, starting with the minimum position/value x, and its m-1 consecutive positions.
- Therefore the 'sum of positions' for each integer 1 to n can be calculated, where sum_of_position(k) = sumcur= Position_of_1 + Position_of_2 + ...Position_of_k.
- Let the minimum element of the above series be x. When the positions are sorted, this will be the first element and the rest will be consecutive.
- Then if the required subarray exists, then sum_of_position(m) has to be x + (x+1) + ..(x+m-1) {m consecutive terms} = x*m - m + m*(m+1)/2 .
- If sum of all positions for integers 1 to m is this sum, then for given m, true is returned, else there is no such sub-array.
Below is the implementation of the above approach.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
// Function that returns true if the
// required subarray exists
// in the given array
bool subArray(ll* arr, ll n, ll m)
{
ll i;
// Map to store the positions of
// each integer in the original
// permutation
unordered_map<ll, ll> mp;
for (i = 0; i < n; i++) {
// To store the address of each
// entry in arr[n] but with
// 1-based indexing
mp[arr[i]] = i + 1;
}
ll sumcur = 0;
// To track minimum position sumcur
// for sum of all positions
// till this position
ll p = INT_MAX;
vector<ll> ans;
for (i = 1; i <= m; i++) {
// Summing up addresses
sumcur += mp[i];
// Tracking minimum address
// encountered till now
p = min(p, mp[i]);
// The sum of the addresses if
// it forms the required subarray
ll val = p * i - i + (i * (i + 1)) / 2;
if (i == m) {
// If current sum of address
// is equal to val
if (val == sumcur) {
return true;
}
else
return false;
}
}
}
// Driver code
int main()
{
ll arr[] = { 4, 5, 1, 3, 2, 6 };
int n = sizeof(arr) / sizeof(int);
ll m = 3;
if (subArray(arr, n, m))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if the
// required subarray exists
// in the given array
static boolean subArray(int[] arr, int n, int m)
{
int i;
// Map to store the positions of
// each integer in the original
// permutation
HashMap<Integer, Integer> mp =
new HashMap<Integer, Integer> ();
for (i = 0; i < n; i++)
{
// To store the address of each
// entry in arr[n] but with
// 1-based indexing
mp.put(arr[i], i + 1);
}
int sumcur = 0;
// To track minimum position sumcur
// for sum of all positions
// tiint this position
int p = Integer.MAX_VALUE;
Vector<Integer> ans = new Vector<Integer>();
for (i = 1; i <= m; i++)
{
// Summing up addresses
sumcur += mp.get(i);
// Tracking minimum address
// encountered tiint now
p = Math.min(p, mp.get(i));
// The sum of the addresses if
// it forms the required subarray
int val = p * i - i + (i * (i + 1)) / 2;
if (i == m)
{
// If current sum of address
// is equal to val
if (val == sumcur)
{
return true;
}
else
return false;
}
}
return false;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4, 5, 1, 3, 2, 6 };
int n = arr.length;
int m = 3;
if (subArray(arr, n, m))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation of the approach
# Function that returns true if the
# required subarray exists
# in the given array
def subArray(arr, n, m):
i = 0
# Map to store the positions of
# each integer in the original
# permutation
mp = dict()
for i in range(n):
# To store the address of each
# entry in arr[n] but with
# 1-based indexing
mp[arr[i]] = i + 1
sumcur = 0
# To track minimum position sumcur
# for sum of all positions
# ti this position
p = 10**9
ans = []
for i in range(1, m + 1):
# Summing up addresses
sumcur += mp[i]
# Tracking minimum address
# encountered ti now
p = min(p, mp[i])
# The sum of the addresses if
# it forms the required subarray
val = p * i - i + (i * (i + 1)) / 2
if (i == m):
# If current sum of address
# is equal to val
if (val == sumcur):
return True
else:
return False
# Driver code
arr = [4, 5, 1, 3, 2, 6]
n = len(arr)
m = 3
if (subArray(arr, n, m)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that returns true if the
// required subarray exists
// in the given array
static bool subArray(int[] arr, int n, int m)
{
int i;
// Map to store the positions of
// each integer in the original
// permutation
Dictionary<int, int> mp =
new Dictionary<int, int> ();
for (i = 0; i < n; i++)
{
// To store the address of each
// entry in arr[n] but with
// 1-based indexing
mp.Add(arr[i], i + 1);
}
int sumcur = 0;
// To track minimum position sumcur
// for sum of all positions
// tiint this position
int p = int.MaxValue;
List<int> ans = new List<int>();
for (i = 1; i <= m; i++)
{
// Summing up addresses
sumcur += mp[i];
// Tracking minimum address
// encountered tiint now
p = Math.Min(p, mp[i]);
// The sum of the addresses if
// it forms the required subarray
int val = p * i - i + (i * (i + 1)) / 2;
if (i == m)
{
// If current sum of address
// is equal to val
if (val == sumcur)
{
return true;
}
else
return false;
}
}
return false;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 4, 5, 1, 3, 2, 6 };
int n = arr.Length;
int m = 3;
if (subArray(arr, n, m))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript implementation of the approach
// Function that returns true if the
// required subarray exists
// in the given array
function subArray(arr, n, m)
{
var i;
// Map to store the positions of
// each integer in the original
// permutation
var mp = new Map();
for(i = 0; i < n; i++)
{
// To store the address of each
// entry in arr[n] but with
// 1-based indexing
mp.set(arr[i], i + 1);
}
var sumcur = 0;
// To track minimum position sumcur
// for sum of all positions
// till this position
var p = 1000000000;
var ans = [];
for(i = 1; i <= m; i++)
{
// Summing up addresses
sumcur += mp.get(i);
// Tracking minimum address
// encountered till now
p = Math.min(p, mp.get(i));
// The sum of the addresses if
// it forms the required subarray
var val = p * i - i +
parseInt((i * (i + 1)) / 2);
if (i == m)
{
// If current sum of address
// is equal to val
if (val == sumcur)
{
return true;
}
else
return false;
}
}
}
// Driver code
var arr = [ 4, 5, 1, 3, 2, 6 ];
var n = arr.length;
var m = 3;
if (subArray(arr, n, m))
document.write("Yes");
else
document.write("No");
// This code is contributed by famously
</script>
Time Complexity: O(N)
Auxiliary Space: O(N) because it is using extra space for map
Another Efficient Approach: Using Sliding Window
We will use a M sized sliding window in which we will keep a count of numbers less than or equal to M and iterate over the array. If the count becomes equal to M, we have found the permutation.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
// Function that returns true if the
// required subarray exists
// in the given array
bool subArray(ll* arr, ll n, ll m)
{
int count = 0;//count less than m
for (int i = 0; i < m; i++){
if (arr[i] <= m)
count++;
}
if (count == m)
return true;
for (int i = m; i < n; i++){
if (arr[i-m] <= m)
count--;
if (arr[i] <= m)
count++;
if (count == m)
return true;
}
return false;
}
// Driver code
int main()
{
ll arr[] = { 4, 5, 1, 3, 2, 6 };
int n = sizeof(arr) / sizeof(int);
ll m = 3;
if (subArray(arr, n, m))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java implementation of the approach
import java.io.*;
class GFG {
// Function that returns true if the
// required subarray exists
// in the given array
static boolean subArray(int[] arr, int n, int m)
{
int count = 0; // count less than m
for (int i = 0; i < m; i++) {
if (arr[i] <= m)
count++;
}
if (count == m)
return true;
for (int i = m; i < n; i++) {
if (arr[i - m] <= m)
count--;
if (arr[i] <= m)
count++;
if (count == m)
return true;
}
return false;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4, 5, 1, 3, 2, 6 };
int n = arr.length;
int m = 3;
if (subArray(arr, n, m))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by subhammahato348.
# Python3 implementation of the approach
# Function that returns true if the
# required subarray exists
# in the given array
def subArray(arr, n, m):
count = 0 # count less than m
for i in range(m):
if (arr[i] <= m):
count += 1
if (count == m):
return True
for i in range(m,n):
if (arr[i-m] <= m):
count -= 1
if (arr[i] <= m):
count += 1
if (count == m):
return True
return False
# Driver code
arr = [ 4, 5, 1, 3, 2, 6 ]
n = len(arr)
m = 3
if (subArray(arr, n, m)):
print("Yes")
else:
print("No")
# This code is contributed by shinjanpatra
// C# implementation of the approach
using System;
class GFG {
// Function that returns true if the
// required subarray exists
// in the given array
static bool subArray(int[] arr, int n, int m)
{
int count = 0; // count less than m
for (int i = 0; i < m; i++) {
if (arr[i] <= m)
count++;
}
if (count == m)
return true;
for (int i = m; i < n; i++) {
if (arr[i - m] <= m)
count--;
if (arr[i] <= m)
count++;
if (count == m)
return true;
}
return false;
}
// Driver code
public static void Main()
{
int[] arr = { 4, 5, 1, 3, 2, 6 };
int n = arr.Length;
int m = 3;
if (subArray(arr, n, m))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by subhammahato348.
<script>
// JavaScript implementation of the approach
// Function that returns true if the
// required subarray exists
// in the given array
function subArray(arr, n, m)
{
let count = 0//count less than m
for (let i = 0; i < m; i++){
if (arr[i] <= m)
count++;
}
if (count == m)
return true;
for (let i = m; i < n; i++){
if (arr[i-m] <= m)
count--;
if (arr[i] <= m)
count++;
if (count == m)
return true;
}
return false;
}
// Driver code
let arr =[ 4, 5, 1, 3, 2, 6 ]
let n = arr.length
let m = 3
if (subArray(arr, n, m))
document.write("Yes")
else
document.write("No")
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(N)
Space Complexity: O(N)
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