Longest subarray with sum not divisible by X

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”.
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 

Output:

3 2 6

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

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My Personal Notes

C++

Java

Python3

C#

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