Check if sum of any subarray is Palindrome or not
Last Updated :
05 Sep, 2022
Given an array arr[] of size N. the task is to check whether there exists any subarray of size atleast 2 such that its sum is palindrome. If such a subarray exists, then print YES. Otherwise, print NO.
Examples:
Input: arr[] = {10, 6, 7, 9, 12}
Output: Yes
Explanation:
The subarray [6, 7, 9] with sum 22 is palindrome.
Input: arr[] = {15, 4, 8, 2}
Output: No
Explanation:
No such subarray exists.
Approach:
To solve the problem follow the steps below:
- Create a prefix sum array of the given array.
- Iterate over the array using nested for loops to denote start and end of subarrays. The sum of the subarray within the indices [x, y] can be obtained by pref[y] - pref[x - 1].
- Check if this sum is Palindrome or not. If any of the sum if palindrome print "Yes", otherwise print "No".
Below is the implementation of above approach:
C++
// C++ program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
#include <bits/stdc++.h>
using namespace std;
// Function which checks whether
// a given number is palindrome or not
bool checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for (int x = n; x != 0; x /= 10) {
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
void findSubarray(int ar[], int n)
{
// Making a prefix sum array of ar[]
int pref[n];
pref[0] = ar[0];
for (int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
bool found = false;
for (int x = 0; x < n; x++) {
for (int y = x + 1; y < n; y++) {
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0) {
sum -= pref[x - 1];
}
if (checkPalindrome(sum)) {
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
// Driver code
int main()
{
int ar[] = { 1, 11, 20, 35 };
int n = sizeof(ar) / sizeof(ar[0]);
findSubarray(ar, n);
return 0;
}
// Java program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
class GFG{
// Function which checks whether
// a given number is palindrome or not
static boolean checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for(int x = n; x != 0; x /= 10)
{
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
static void findSubarray(int []ar, int n)
{
// Making a prefix sum array of ar[]
int []pref = new int[n];
pref[0] = ar[0];
for(int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
boolean found = false;
for(int x = 0; x < n; x++)
{
for(int y = x + 1; y < n; y++)
{
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0)
{
sum -= pref[x - 1];
}
if (checkPalindrome(sum))
{
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
System.out.println("Yes");
else
System.out.println("No");
}
// Driver code
public static void main(String args[])
{
int []ar = { 1, 11, 20, 35 };
int n = ar.length;
findSubarray(ar, n);
}
}
// This code is contributed by AnkitRai01
# Python3 program to check if sum of
# any subarray of size atleast 2 is
# palindrome or not
# Function which checks whether a
# given number is palindrome or not
def checkPalindrome(n):
# Store the reverse
# of the number n
rev = 0
x = n
while(x != 0):
d = x % 10
rev = rev * 10 + d
x = x // 10
if (rev == n):
return True
else:
return False
# Function which checks if the
# requires subarray exists or not
def findSubarray(ar, n):
# Making a prefix sum array of ar[]
pref = [0 for i in range(n)]
pref[0] = ar[0]
for x in range(1, n):
pref[x] = pref[x - 1] + ar[x]
# Boolean variable that will store
# whether such subarray exists or not
found = False
for x in range(n):
for y in range(x + 1, n, 1):
# Sum stores the sum of subarray
# from index x to y of array
sum = pref[y]
if (x > 0):
sum -= pref[x - 1]
if (checkPalindrome(sum)):
# Required subarray is found
found = True
break
if (found):
break
if (found):
print("Yes")
else:
print("No")
# Driver code
if __name__ == '__main__':
ar = [ 1, 11, 20, 35 ]
n = len(ar)
findSubarray(ar, n)
# This code is contributed by Surendra_Gangwar
// C# program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
using System;
class GFG{
// Function which checks whether
// a given number is palindrome or not
static bool checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for(int x = n; x != 0; x /= 10)
{
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
static void findSubarray(int []ar, int n)
{
// Making a prefix sum array of ar[]
int []pref = new int[n];
pref[0] = ar[0];
for(int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
bool found = false;
for(int x = 0; x < n; x++)
{
for(int y = x + 1; y < n; y++)
{
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0)
{
sum -= pref[x - 1];
}
if (checkPalindrome(sum))
{
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver code
public static void Main()
{
int []ar = { 1, 11, 20, 35 };
int n = ar.Length;
findSubarray(ar, n);
}
}
// This code is contributed by Code_Mech
<script>
// JavaScript program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
// Function which checks whether
// a given number is palindrome or not
function checkPalindrome(n)
{
// Store the reverse of
// the number n
var rev = 0;
for (var x = n; x != 0; x = parseInt(x/10)) {
var d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
function findSubarray(ar, n)
{
// Making a prefix sum array of ar[]
var pref = Array(n).fill(0);
pref[0] = ar[0];
for (var x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
var found = false;
for (var x = 0; x < n; x++) {
for (var y = x + 1; y < n; y++) {
// sum stores the sum of subarray
// from index x to y of array
var sum = pref[y];
if (x > 0) {
sum -= pref[x - 1];
}
if (checkPalindrome(sum)) {
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
document.write( "Yes" );
else
document.write( "No" );
}
// Driver code
var ar = [1, 11, 20, 35];
var n = ar.length;
findSubarray(ar, n);
</script>
Time Complexity: O(n2logn)
Auxiliary Space: O(n) because using extra space for pref array
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