Length of longest subarray whose sum is not divisible by integer K

Given an array arr[] of size N and an integer k, our task is to find the length of longest subarray whose sum of elements is not divisible by k. If no such subarray exists then return -1.

Examples:

Input: arr[] = {8, 4, 3, 1, 5, 9, 2}, k = 2
Output: 5
Explanation:
The subarray is {8, 4, 3, 1, 5} with sum = 21, is not divisible by 2.

Input: arr[] = {6, 3, 12, 15}, k = 3
Output: -1
Explanation:
There is no subarray which is not divisible by 3.

Naive Approach: The idea is to consider all the subarrays and return the length of the longest subarray such that the sum of its elements is not divisible by k.



Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The main observation is that removing an element that is divisible by k will not contribute to the solution, but if we remove an element that is not divisible by k then the sum would not be divisible by k.

  • Therefore, let the leftmost non-multiple of k be at index left, and the rightmost non-multiple of k be at index right.
  • Remove either the prefix elements up to index left, or the suffix element up to index right and remove the elements which have lesser number of elements.
  • There are two corner cases in this problem. First is, if every element of the array are divisible by k, then no such subarray exist so return -1. Secondly, if some of the whole array is not divisible by k, then the subarray will be the array itself, so return the size of the array.

Below is the implementation of the above approach:

C++

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// C++ Program to find the length of
// the longest subarray whose sum is
// not divisible by integer K
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the longest subarray
// with sum is not divisible by k
int MaxSubarrayLength(int arr[], int n, int k)
{
    // left is the index of the
    // leftmost element that is
    // not divisible by k
    int left = -1;
  
    // right is the index of the
    // rightmost element that is
    // not divisible by k
    int right;
  
    // sum of the array
    int sum = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Find the element that
        // is not multiple of k
        if ((arr[i] % k) != 0) {
  
            // left = -1 means we are
            // finding the leftmost
            // element that is not
            // divisible by k
            if (left == -1) {
                left = i;
            }
  
            // Updating the
            // rightmost element
            right = i;
        }
  
        // update the sum of the
        // array up to the index i
        sum += arr[i];
    }
  
    // Check if the sum of the
    // array is not divisible
    // by k, then return the
    // size of array
    if ((sum % k) != 0) {
        return n;
    }
  
    // All elements of array
    // are divisible by k,
    // then no such subarray
    // possible so return -1
    else if (left == -1) {
        return -1;
    }
  
    else {
        // length of prefix elements
        // that can be removed
        int prefix_length = left + 1;
  
        // length of suffix elements
        // that can be removed
        int suffix_length = n - right;
  
        // Return the length of
        // subarray after removing
        // the elements which have
        // lesser number of elements
        return n - min(prefix_length,
                       suffix_length);
    }
}
  
// Driver Code
int main()
{
  
    int arr[] = { 6, 3, 12, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
  
    cout << MaxSubarrayLength(arr, n, K);
  
    return 0;
}

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Java

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// Java program to find the length of
// the longest subarray whose sum is
// not divisible by integer K
import java.util.*;
  
class GFG{
  
// Function to find the longest subarray
// with sum is not divisible by k
static int MaxSubarrayLength(int arr[], int n,
                                        int k)
{
      
    // left is the index of the
    // leftmost element that is
    // not divisible by k
    int left = -1;
  
    // right is the index of the
    // rightmost element that is
    // not divisible by k
    int right = 0;
  
    // sum of the array
    int sum = 0;
  
    for(int i = 0; i < n; i++)
    {
          
        // Find the element that
        // is not multiple of k
        if ((arr[i] % k) != 0
        {
  
            // left = -1 means we are
            // finding the leftmost
            // element that is not
            // divisible by k
            if (left == -1
            {
                left = i;
            }
  
            // Updating the
            // rightmost element
            right = i;
        }
  
        // Update the sum of the
        // array up to the index i
        sum += arr[i];
    }
  
    // Check if the sum of the
    // array is not divisible
    // by k, then return the
    // size of array
    if ((sum % k) != 0)
    {
        return n;
    }
  
    // All elements of array
    // are divisible by k,
    // then no such subarray
    // possible so return -1
    else if (left == -1)
    {
        return -1;
    }
    else 
    {
          
        // Length of prefix elements
        // that can be removed
        int prefix_length = left + 1;
  
        // Length of suffix elements
        // that can be removed
        int suffix_length = n - right;
  
        // Return the length of
        // subarray after removing
        // the elements which have
        // lesser number of elements
        return n - Math.min(prefix_length,
                            suffix_length);
    }
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 6, 3, 12, 15 };
    int n = arr.length;
    int K = 3;
      
    System.out.println(MaxSubarrayLength(arr, n, K));
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program to find the length of
# the longest subarray whose sum is
# not divisible by integer 
  
# Function to find the longest subarray
# with sum is not divisible by k
def MaxSubarrayLength(arr, n, k):
  
    # left is the index of the
    # leftmost element that is
    # not divisible by k
    left = -1
  
    # sum of the array
    sum = 0
  
    for i in range(n):
  
        # Find the element that
        # is not multiple of k
        if ((arr[i] % k) != 0):
  
            # left = -1 means we are
            # finding the leftmost
            # element that is not
            # divisible by k
            if (left == -1):
                left = i
  
            # Updating the
            # rightmost element
            right = i
  
        # Update the sum of the
        # array up to the index i
        sum += arr[i]
  
    # Check if the sum of the
    # array is not divisible
    # by k, then return the
    # size of array
    if ((sum % k) != 0):
        return n
  
    # All elements of array
    # are divisible by k,
    # then no such subarray
    # possible so return -1
    elif(left == -1):
        return -1
  
    else:
          
        # length of prefix elements
        # that can be removed
        prefix_length = left + 1
  
        # length of suffix elements
        # that can be removed
        suffix_length = n - right
  
        # Return the length of
        # subarray after removing
        # the elements which have
        # lesser number of elements
        return n - min(prefix_length,
                       suffix_length)
                         
# Driver Code
if __name__ == "__main__":
  
    arr = [ 6, 3, 12, 15 ]
    n = len(arr)
    K = 3
  
    print(MaxSubarrayLength(arr, n, K))
  
# This code is contributed by chitranayal

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C#

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// C# program to find the length of
// the longest subarray whose sum is
// not divisible by integer K
using System;
  
class GFG{
  
// Function to find the longest subarray
// with sum is not divisible by k
static int MaxSubarrayLength(int []arr, int n,
                                        int k)
{
      
    // left is the index of the
    // leftmost element that is
    // not divisible by k
    int left = -1;
  
    // right is the index of the
    // rightmost element that is
    // not divisible by k
    int right = 0;
  
    // sum of the array
    int sum = 0;
  
    for(int i = 0; i < n; i++)
    {
          
        // Find the element that
        // is not multiple of k
        if ((arr[i] % k) != 0) 
        {
  
            // left = -1 means we are
            // finding the leftmost
            // element that is not
            // divisible by k
            if (left == -1) 
            {
                left = i;
            }
  
            // Updating the
            // rightmost element
            right = i;
        }
  
        // Update the sum of the
        // array up to the index i
        sum += arr[i];
    }
  
    // Check if the sum of the
    // array is not divisible
    // by k, then return the
    // size of array
    if ((sum % k) != 0)
    {
        return n;
    }
  
    // All elements of array
    // are divisible by k,
    // then no such subarray
    // possible so return -1
    else if (left == -1)
    {
        return -1;
    }
    else
    {
          
        // Length of prefix elements
        // that can be removed
        int prefix_length = left + 1;
  
        // Length of suffix elements
        // that can be removed
        int suffix_length = n - right;
  
        // Return the length of
        // subarray after removing
        // the elements which have
        // lesser number of elements
        return n - Math.Min(prefix_length,
                            suffix_length);
    }
}
  
// Driver code
public static void Main(string[] args)
{
    int []arr = { 6, 3, 12, 15 };
    int n = arr.Length;
    int K = 3;
      
    Console.Write(MaxSubarrayLength(arr, n, K));
}
}
  
// This code is contributed by rutvik_56

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Output:

-1

Time Complexity: O(N)
Auxiliary Space: O(1)

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Improved By : chitranayal, offbeat, rutvik_56