Longest subarray with sum not divisible by X

Last Updated : 23 Mar, 2023

Given an array arr[] and an integer X, the task is to print the longest subarray such that the sum of its elements isn’t divisible by X. If no such subarray exists, print “-1”.
Note: If more than one subarray exists with the given property, print any one of them.
Examples:

Input: arr[] = {1, 2, 3} X = 3
Output: 2 3
Explanation:
The subarray {2, 3} has a sum of elements 5, which isn’t divisible by 3.
Input: arr[] = {2, 6} X = 2
Output: -1
Explanation:
All possible subarrays {1}, {2}, {1, 2} have an even sum.
Therefore, the answer is -1.

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays and keep calculating its sum. If any subarray is found to have sum not divisible by X, compare the length with maximum length obtained(maxm) and update the maxm accordingly and update the starting index and ending index of the subarray. Finally, print the subarray having the stored starting and ending indices. If there is no such subarray then print “-1”.

C++

   // C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the longest
// subarray with sum of elements
// not divisible by X
void max_length(int n, int x,vector<int> a)
{
    // Variable to store start and end index
    int maxm = -1, start = -1, end = -1;
 
    // traversing to generate all subarray
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            // variable to store sum
            int sum = 0;
            for (int k = i; k <= j; k++) {
                sum += a[k];
            }
            // Checking if sum is divisible by x
            // or not. If not then update the length 
            // if it greater than all previous length
            if (sum % x != 0 && j - i + 1 > maxm) {
                maxm = j - i + 1;
                start = i;
                end = j;
            }
        }
    }
     
    // If there is no such subarray then print “-1”
    if (maxm == -1) {
        cout << "-1\n";
    } 
    // print the subarray having the stored starting and ending indices
    else {
        for (int i = start; i <= end; i++) {
            cout << a[i] << " ";
        }
        cout << "\n";
    }
 
}
 
// Driver Code
int main()
{
    int x = 3;
  
    vector<int> v = { 1, 3, 2, 6 };
    int N = v.size();
  
    max_length(N, x, v);
  
    return 0;
}
 
// This code is contributed by Pushpesh Raj.
 
  

Java

   // Java Program to implement
// the above approach
import java.util.*;
 
class Main {
    // Function to print the longest
    // subarray with sum of elements
    // not divisible by X
    static void max_length(int n, int x,
                           ArrayList<Integer> a)
    {
        // Variable to store start and end index
        int maxm = -1, start = -1, end = -1;
 
        // traversing to generate all subarray
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                // variable to store sum
                int sum = 0;
                for (int k = i; k <= j; k++) {
                    sum += a.get(k);
                }
                // Checking if sum is divisible by x
                // or not. If not then update the length
                // if it greater than all previous length
                if (sum % x != 0 && j - i + 1 > maxm) {
                    maxm = j - i + 1;
                    start = i;
                    end = j;
                }
            }
        }
 
        // If there is no such subarray then print “-1”
        if (maxm == -1) {
            System.out.println("-1");
        }
        // print the subarray having the stored starting and
        // ending indices
        else {
            for (int i = start; i <= end; i++) {
                System.out.print(a.get(i) + " ");
            }
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int x = 3;
 
        ArrayList<Integer> v = new ArrayList<Integer>(
            Arrays.asList(1, 3, 2, 6));
        int N = v.size();
 
        max_length(N, x, v);
    }
}
 
  

Python3

   # Python Program to implement
# the above approach
 
def max_length(n, x, a):
    # Variable to store start and end index
    maxm = -1
    start = -1
    end = -1
 
    # traversing to generate all subarray
    for i in range(0, n):
        for j in range(i, n):
 
            # variable to store sum
            sum1 = 0
            for k in range(i, j + 1):
                sum1 += a[k]
             
            # Checking if sum is divisible by x
            # or not. If not then update the length 
            # if it greater than all previous length
            if sum1 % x != 0 and j - i + 1 > maxm:
                maxm = j - i + 1
                start = i
                end = j
     
    # If there is no such subarray then print “-1”
    if maxm == -1:
        print("-1")
    # print the subarray having the stored starting and ending indices
    else:
        for i in range(start, end + 1):
            print(a[i], end=" ")
        print()
 
 
# Driver Code
if __name__ == "__main__":
    x = 3
 
    v = [1, 3, 2, 6]
    N = len(v)
 
    max_length(N, x, v)
 
  

C#

   // C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG {
    // Function to print the longest
    // subarray with sum of elements
    // not divisible by X
    static void max_length(int n, int x, List<int> a)
    {
        // Variable to store start and end index
        int maxm = -1, start = -1, end = -1;
 
        // traversing to generate all subarray
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                // variable to store sum
                int sum = 0;
                for (int k = i; k <= j; k++) {
                    sum += a[k];
                }
                // Checking if sum is divisible by x
                // or not. If not then update the length
                // if it greater than all previous length
                if (sum % x != 0 && j - i + 1 > maxm) {
                    maxm = j - i + 1;
                    start = i;
                    end = j;
                }
            }
        }
 
        // If there is no such subarray then print “-1”
        if (maxm == -1) {
            Console.WriteLine("-1");
        }
        // print the subarray having the stored starting and
        // ending indices
        else {
            for (int i = start; i <= end; i++) {
                Console.Write(a[i] + " ");
            }
            Console.WriteLine();
        }
    }
 
    // Driver Code
    static void Main(string[] args)
    {
        int x = 3;
 
        List<int> v = new List<int>{ 1, 3, 2, 6 };
        int N = v.Count;
 
        max_length(N, x, v);
    }
}
 
  

Javascript

   // JavaScript program to implement
// the above approach
 
function max_length(n, x, a) {
  // Variable to store start and end index
  let maxm = -1;
  let start = -1;
  let end = -1;
 
  // traversing to generate all subarray
  for (let i = 0; i < n; i++) {
    for (let j = i; j < n; j++) {
 
      // variable to store sum
      let sum1 = 0;
      for (let k = i; k <= j; k++) {
        sum1 += a[k];
      }
 
      // Checking if sum is divisible by x
      // or not. If not then update the length 
      // if it greater than all previous length
      if (sum1 % x !== 0 && j - i + 1 > maxm) {
        maxm = j - i + 1;
        start = i;
        end = j;
      }
    }
  }
 
  // If there is no such subarray then print “-1”
  if (maxm === -1) {
    console.log("-1");
  } 
  // print the subarray having the stored starting and ending indices
  else { temp=""
    for (let i = start; i <= end; i++) {
      temp = temp + a[i]+" ";
    }
    console.log(temp);
  }
}
 
// Driver Code
let x = 3;
 
let v = [1, 3, 2, 6];
let N = v.length;
 
max_length(N, x, v);
 
  

Output

3 2 6

Time Complexity: O(N²)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach we will find the prefix and suffix array sum. Follow the steps below:

Generate the prefix sum array and suffix sum array.
Iterate from [0, N – 1] using Two Pointers and choose the prefix and suffix sum of the element at each index which is not divisible by X. Store the starting index and ending index of the subarray.
After completing the above steps, if there exist a subarray with sum not divisible by X, then print the subarray having the stored starting and ending indices.
If there is no such subarray then print “-1”.

Below is the implementation of the above approach:

C++

   #include <iostream>
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the longest
// subarray with sum of elements
// not divisible by X
void max_length(int N, int x,
                vector<int>& v)
{
    int i, a;
 
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    vector<int> preff, suff;
    int ct = 0;
 
    for (i = 0; i < N; i++) {
 
        a = v[i];
 
        // If array element is
        // divisibile by x
        if (a % x == 0) {
 
            // Increase count
            ct += 1;
        }
    }
 
    // If all the array elements
    // are divisible by x
    if (ct == N) {
 
        // No subarray possible
        cout << -1 << endl;
        return;
    }
 
    // Reverse v to calculate the
    // suffix sum
    reverse(v.begin(), v.end());
 
    suff.push_back(v[0]);
 
    // Calculate the suffix sum
    for (i = 1; i < N; i++) {
        suff.push_back(v[i]
                       + suff[i - 1]);
    }
 
    // Reverse to original form
    reverse(v.begin(), v.end());
 
    // Reverse the suffix sum array
    reverse(suff.begin(), suff.end());
 
    preff.push_back(v[0]);
 
    // Calculate the prefix sum
    for (i = 1; i < N; i++) {
        preff.push_back(v[i]
                        + preff[i - 1]);
    }
 
    int ans = 0;
 
    // Stores the starting index
    // of required subarray
    int lp = 0;
 
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
 
    for (i = 0; i < N; i++) {
 
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff[i] % x != 0
            && (ans < (N - 1))) {
 
            lp = i;
            rp = N - 1;
 
            // Update the answer
            ans = max(ans, N - i);
        }
 
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff[i] % x != 0
            && (ans < (i + 1))) {
 
            lp = 0;
            rp = i;
 
            // Update the answer
            ans = max(ans, i + 1);
        }
    }
 
    // Print the longest subarray
    for (i = lp; i <= rp; i++) {
        cout << v[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int x = 3;
 
    vector<int> v = { 1, 3, 2, 6 };
    int N = v.size();
 
    max_length(N, x, v);
 
    return 0;
}
 
  

Java

   // Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to print the longest
// subarray with sum of elements
// not divisible by X
static void max_length(int N, int x,
                       int []v)
{
    int i, a;
 
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    List<Integer> preff = new Vector<Integer>();
    List<Integer> suff = new Vector<Integer>();
     
    int ct = 0;
 
    for(i = 0; i < N; i++) 
    {
        a = v[i];
 
        // If array element is
        // divisibile by x
        if (a % x == 0)
        {
             
            // Increase count
            ct += 1;
        }
    }
 
    // If all the array elements
    // are divisible by x
    if (ct == N) 
    {
         
        // No subarray possible
        System.out.print(-1 + "\n");
        return;
    }
 
    // Reverse v to calculate the
    // suffix sum
    v = reverse(v);
 
    suff.add(v[0]);
 
    // Calculate the suffix sum
    for(i = 1; i < N; i++)
    {
        suff.add(v[i] + suff.get(i - 1));
    }
 
    // Reverse to original form
    v = reverse(v);
 
    // Reverse the suffix sum array
    Collections.reverse(suff);
 
    preff.add(v[0]);
 
    // Calculate the prefix sum
    for(i = 1; i < N; i++)
    {
        preff.add(v[i] + preff.get(i - 1));
    }
 
    int ans = 0;
 
    // Stores the starting index
    // of required subarray
    int lp = 0;
 
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
 
    for(i = 0; i < N; i++)
    {
         
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff.get(i) % x != 0 &&
           (ans < (N - 1))) 
        {
            lp = i;
            rp = N - 1;
 
            // Update the answer
            ans = Math.max(ans, N - i);
        }
 
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff.get(i) % x != 0 &&
           (ans < (i + 1)))
        {
            lp = 0;
            rp = i;
 
            // Update the answer
            ans = Math.max(ans, i + 1);
        }
    }
 
    // Print the longest subarray
    for(i = lp; i <= rp; i++)
    {
        System.out.print(v[i] + " ");
    }
}
 
static int[] reverse(int a[]) 
{
    int i, n = a.length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Driver Code
public static void main(String[] args)
{
    int x = 3;
    int []v = { 1, 3, 2, 6 };
    int N = v.length;
 
    max_length(N, x, v);
}
}
 
// This code is contributed by PrinciRaj1992
 
  

Python3

   # Python3 program to implement 
# the above approach 
 
# Function to print the longest 
# subarray with sum of elements 
# not divisible by X 
def max_length(N, x, v):
     
    # Pref[] stores the prefix sum 
    # Suff[] stores the suffix sum 
    preff, suff = [], []
    ct = 0
     
    for i in range(N):
        a = v[i]
         
        # If array element is 
        # divisibile by x 
        if a % x == 0:
             
            # Increase count 
            ct += 1
             
    # If all the array elements 
    # are divisible by x 
    if ct == N:
         
        # No subarray possible 
        print(-1)
        return
     
    # Reverse v to calculate the 
    # suffix sum 
    v.reverse()
     
    suff.append(v[0])
     
    # Calculate the suffix sum 
    for i in range(1, N):
        suff.append(v[i] + suff[i - 1])
         
    # Reverse to original form 
    v.reverse()
     
    # Reverse the suffix sum array
    suff.reverse()
     
    preff.append(v[0])
     
    # Calculate the prefix sum
    for i in range(1, N):
        preff.append(v[i] + preff[i - 1])
         
    ans = 0
     
    # Stores the starting index 
    # of required subarray 
    lp = 0
     
    # Stores the ending index 
    # of required subarray 
    rp = N - 1
     
    for i in range(N):
         
        # If suffix sum till i-th 
        # index is not divisible by x 
        if suff[i] % x != 0 and ans < N - 1:
            lp = i
            rp = N - 1
             
            # Update the answer
            ans = max(ans, N - i)
             
        # If prefix sum till i-th 
        # index is not divisible by x 
        if preff[i] % x != 0 and ans < i + 1:
            lp = 0
            rp = i
             
            # Update the answer
            ans = max(ans, i + 1)
             
    # Print the longest subarray
    for i in range(lp, rp + 1):
        print(v[i], end = " ")
         
# Driver code
x = 3
v = [ 1, 3, 2, 6 ]
N = len(v)
 
max_length(N, x, v)
 
# This code is contributed by Stuti Pathak
 
  

C#

   // C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to print the longest
// subarray with sum of elements
// not divisible by X
static void max_length(int N, int x,
                       int []v)
{
    int i, a;
 
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    List<int> preff = new List<int>();
    List<int> suff = new List<int>();
     
    int ct = 0;
 
    for(i = 0; i < N; i++) 
    {
        a = v[i];
 
        // If array element is
        // divisibile by x
        if (a % x == 0)
        {
             
            // Increase count
            ct += 1;
        }
    }
 
    // If all the array elements
    // are divisible by x
    if (ct == N) 
    {
         
        // No subarray possible
        Console.Write(-1 + "\n");
        return;
    }
 
    // Reverse v to calculate the
    // suffix sum
    v = reverse(v);
 
    suff.Add(v[0]);
 
    // Calculate the suffix sum
    for(i = 1; i < N; i++)
    {
        suff.Add(v[i] + suff[i - 1]);
    }
 
    // Reverse to original form
    v = reverse(v);
 
    // Reverse the suffix sum array
    suff.Reverse();
 
    preff.Add(v[0]);
 
    // Calculate the prefix sum
    for(i = 1; i < N; i++)
    {
        preff.Add(v[i] + preff[i - 1]);
    }
 
    int ans = 0;
 
    // Stores the starting index
    // of required subarray
    int lp = 0;
 
    // Stores the ending index
    // of required subarray
    int rp = N - 1;
 
    for(i = 0; i < N; i++)
    {
         
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff[i] % x != 0 &&
               (ans < (N - 1))) 
        {
            lp = i;
            rp = N - 1;
 
            // Update the answer
            ans = Math.Max(ans, N - i);
        }
 
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff[i] % x != 0 &&
                (ans < (i + 1)))
        {
            lp = 0;
            rp = i;
 
            // Update the answer
            ans = Math.Max(ans, i + 1);
        }
    }
 
    // Print the longest subarray
    for(i = lp; i <= rp; i++)
    {
        Console.Write(v[i] + " ");
    }
}
 
static int[] reverse(int []a) 
{
    int i, n = a.Length, t;
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Driver Code
public static void Main(String[] args)
{
    int x = 3;
    int []v = { 1, 3, 2, 6 };
    int N = v.Length;
 
    max_length(N, x, v);
}
}
 
// This code is contributed by PrinciRaj1992
 
  

Javascript

   // JS Program to implement
// the above approach
 
// Function to print the longest
// subarray with sum of elements
// not divisible by X
function max_length( N,  x,
                v)
{
    let i, a;
 
    // Pref[] stores the prefix sum
    // Suff[] stores the suffix sum
    let preff = [], suff = [];
    let ct = 0;
 
    for (i = 0; i < N; i++) {
 
        a = v[i];
 
        // If array element is
        // divisibile by x
        if (a % x == 0) {
 
            // Increase count
            ct += 1;
        }
    }
 
    // If all the array elements
    // are divisible by x
    if (ct == N) {
 
        // No subarray possible
        console.log(-1)
        return;
    }
 
    // Reverse v to calculate the
    // suffix sum
    v.reverse()
 
    suff.push(v[0]);
 
    // Calculate the suffix sum
    for (i = 1; i < N; i++) {
        suff.push(v[i]
                       + suff[i - 1]);
    }
 
    // Reverse to original form
    v.reverse()
 
    // Reverse the suffix sum array
    suff.reverse()
 
    preff.push(v[0]);
 
    // Calculate the prefix sum
    for (i = 1; i < N; i++) {
        preff.push(v[i]
                        + preff[i - 1]);
    }
 
    let ans = 0;
 
    // Stores the starting index
    // of required subarray
    let lp = 0;
 
    // Stores the ending index
    // of required subarray
    let rp = N - 1;
 
    for (i = 0; i < N; i++) {
 
        // If suffix sum till i-th
        // index is not divisible by x
        if (suff[i] % x != 0
            && (ans < (N - 1))) {
 
            lp = i;
            rp = N - 1;
 
            // Update the answer
            ans = Math.max(ans, N - i);
        }
 
        // If prefix sum till i-th
        // index is not divisible by x
        if (preff[i] % x != 0
            && (ans < (i + 1))) {
 
            lp = 0;
            rp = i;
 
            // Update the answer
            ans = Math.max(ans, i + 1);
        }
    }
 
    // Print the longest subarray
    for (i = lp; i <= rp; i++) {
       process.stdout.write(v[i] + " ");
    }
}
 
// Driver Code
let x = 3;
 
let v = [ 1, 3, 2, 6 ];
let N = v.length;
 
max_length(N, x, v);
 
// This code is contributed by phasing17
 
  

Output:

3 2 6

Time Complexity: O(N)
Auxiliary Space: O(N)

Count of longest possible subarrays with sum not divisible by K

RishavSinghMehta

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Longest subarray with sum not divisible by X

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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