Longest subarray with sum divisible by k

Difficulty Level : Hard
Last Updated : 08 Sep, 2020

Given an arr[] containing n integers and a positive integer k. The problem is to find the length of the longest subarray with sum of the elements divisible by the given value k.
Examples:

Input : arr[] = {2, 7, 6, 1, 4, 5}, k = 3
Output : 4
The subarray is {7, 6, 1, 4} with sum 18,
which is divisible by 3.

Input : arr[] = {-2, 2, -5, 12, -11, -1, 7}
Output : 5

Method 1 (Naive Approach): Consider all the subarrays and return the length of the subarray with sum divisible by k and has the longest length.
Time Complexity: O(n²).

Method 2 (Efficient Approach): Create an array mod_arr[] where mod_arr[i] stores (sum(arr[0]+arr[1]..+arr[i]) % k). Create a hash table having tuple as (ele, idx), where ele represents an element of mod_arr[] and idx represents the element’s index of first occurrence in mod_arr[]. Now, traverse mod_arr[] from i = 0 to n and follow the steps given below.

If mod_arr[i] == 0, then update maxLen = (i + 1).
Else if mod_arr[i] is not present in the hash table, then create tuple (mod_arr[i], i) in the hash table.
Else, get the value associated with mod_arr[i] in the hash table. Let this be idx.
If maxLen < (i – idx), then update maxLen = (i – idx).

Finally return maxLen.

C++

// C++ implementation to find the longest subarray 
// with sum divisible by k 
#include <bits/stdc++.h> 
  
using namespace std; 
  
// function to find the longest subarray 
// with sum divisible by k 
int longSubarrWthSumDivByK(int arr[],  
                          int n, int k) 
{ 
    // unodered map 'um' implemented as 
    // hash table 
    unordered_map<int, int> um; 
      
    // 'mod_arr[i]' stores (sum[0..i] % k) 
    int mod_arr[n], max = 0; 
    int curr_sum = 0; 
      
    // traverse arr[] and build up the 
    // array 'mod_arr[]' 
    for (int i = 0; i < n; i++) 
    { 
        curr_sum += arr[i]; 
          
        // as the sum can be negative, taking modulo twice 
        mod_arr[i] = ((curr_sum % k) + k) % k;         
    }     
      
    for (int i = 0; i < n; i++) 
    { 
        // if true then sum(0..i) is divisible 
        // by k 
        if (mod_arr[i] == 0) 
            // update 'max' 
            max = i + 1; 
          
        // if value 'mod_arr[i]' not present in 'um' 
        // then store it in 'um' with index of its 
        // first occurrence         
        else if (um.find(mod_arr[i]) == um.end()) 
            um[mod_arr[i]] = i; 
              
        else
            // if true, then update 'max' 
            if (max < (i - um[mod_arr[i]])) 
                max = i - um[mod_arr[i]];             
    } 
      
    // required length of longest subarray with 
    // sum divisible by 'k' 
    return max; 
}                           
  
// Driver program to test above 
int main() 
{ 
    int arr[] = {2, 7, 6, 1, 4, 5}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int k = 3; 
      
    cout << "Length = "
         << longSubarrWthSumDivByK(arr, n, k); 
           
    return 0;      
} 

Java

// Java implementation to find the longest  
// subarray with sum divisible by k 
import java.io.*; 
import java.util.*; 
  
class GfG { 
          
    // function to find the longest subarray 
    // with sum divisible by k 
    static int longSubarrWthSumDivByK(int arr[],  
                                      int n, int k) 
    { 
        // unodered map 'um' implemented as 
        // hash table 
        HashMap<Integer, Integer> um= new HashMap<Integer, Integer>(); 
          
        // 'mod_arr[i]' stores (sum[0..i] % k) 
        int mod_arr[]= new int[n]; 
        int max = 0; 
        int curr_sum = 0; 
          
        // traverse arr[] and build up the 
        // array 'mod_arr[]' 
        for (int i = 0; i < n; i++) 
        { 
            curr_sum += arr[i]; 
              
            // as the sum can be negative,  
            // taking modulo twice 
            mod_arr[i] = ((curr_sum % k) + k) % k;      
        }  
          
        for (int i = 0; i < n; i++) 
        { 
            // if true then sum(0..i) is  
            // divisible by k 
            if (mod_arr[i] == 0) 
                // update 'max' 
                max = i + 1; 
              
            // if value 'mod_arr[i]' not present in 'um' 
            // then store it in 'um' with index of its 
            // first occurrence      
            else if (um.containsKey(mod_arr[i]) == false) 
                um.put(mod_arr[i] , i); 
                  
            else
                // if true, then update 'max' 
                if (max < (i - um.get(mod_arr[i]))) 
                    max = i - um.get(mod_arr[i]);          
        } 
          
        // required length of longest subarray with 
        // sum divisible by 'k' 
        return max; 
    }     
      
    public static void main (String[] args)  
    { 
        int arr[] = {2, 7, 6, 1, 4, 5}; 
        int n = arr.length; 
        int k = 3; 
          
        System.out.println("Length = "+  
                            longSubarrWthSumDivByK(arr, n, k)); 
          
    } 
} 
  
// This code is contributed by Gitanjali. 

Python3

# Python3 implementation to find the  
# longest subarray with sum divisible by k 
  
# Function to find the longest 
# subarray with sum divisible by k 
def longSubarrWthSumDivByK(arr, n, k): 
      
    # unodered map 'um' implemented  
    # as hash table 
    um = {} 
  
    # 'mod_arr[i]' stores (sum[0..i] % k) 
    mod_arr = [0 for i in range(n)] 
    max = 0
    curr_sum = 0
      
    # Traverse arr[] and build up  
    # the array 'mod_arr[]' 
    for i in range(n): 
        curr_sum += arr[i] 
          
        # As the sum can be negative, 
        # taking modulo twice 
        mod_arr[i] = ((curr_sum % k) + k) % k  
      
    for i in range(n): 
          
        # If true then sum(0..i) is  
        # divisible by k 
        if (mod_arr[i] == 0): 
              
            # Update 'max' 
            max = i + 1
          
        # If value 'mod_arr[i]' not present in  
        # 'um' then store it in 'um' with index  
        # of its first occurrence  
        elif (mod_arr[i] not in um): 
            um[mod_arr[i]] = i 
              
        else: 
              
            # If true, then update 'max' 
            if (max < (i - um[mod_arr[i]])): 
                max = i - um[mod_arr[i]]          
      
    # Required length of longest subarray  
    # with sum divisible by 'k' 
    return max    
  
# Driver Code 
if __name__ == '__main__': 
      
    arr = [ 2, 7, 6, 1, 4, 5 ] 
    n = len(arr) 
    k = 3
      
    print("Length =",  
           longSubarrWthSumDivByK(arr, n, k)) 
      
# This code is contributed by 
# Surendra_Gangwar 

C#

using System; 
using System.Collections.Generic; 
  
// C# implementation to find the longest   
// subarray with sum divisible by k  
  
public class GfG 
{ 
  
    // function to find the longest subarray  
    // with sum divisible by k  
    public static int longSubarrWthSumDivByK(int[] arr, int n, int k) 
    { 
        // unodered map 'um' implemented as  
        // hash table  
        Dictionary<int, int> um = new Dictionary<int, int>(); 
  
        // 'mod_arr[i]' stores (sum[0..i] % k)  
        int[] mod_arr = new int[n]; 
        int max = 0; 
        int curr_sum = 0; 
  
        // traverse arr[] and build up the  
        // array 'mod_arr[]'  
        for (int i = 0; i < n; i++) 
        { 
            curr_sum += arr[i]; 
  
            // as the sum can be negative,   
            // taking modulo twice  
            mod_arr[i] = ((curr_sum % k) + k) % k; 
        } 
  
        for (int i = 0; i < n; i++) 
        { 
            // if true then sum(0..i) is   
            // divisible by k  
            if (mod_arr[i] == 0) 
            { 
                // update 'max'  
                max = i + 1; 
            } 
  
            // if value 'mod_arr[i]' not present in 'um'  
            // then store it in 'um' with index of its  
            // first occurrence       
            else if (um.ContainsKey(mod_arr[i]) == false) 
            { 
                um[mod_arr[i]] = i; 
            } 
  
            else
            { 
                // if true, then update 'max'  
                if (max < (i - um[mod_arr[i]])) 
                { 
                    max = i - um[mod_arr[i]]; 
                } 
            } 
        } 
  
        // required length of longest subarray with  
        // sum divisible by 'k'  
        return max; 
    } 
  
    public static void Main(string[] args) 
    { 
        int[] arr = new int[] {2, 7, 6, 1, 4, 5}; 
        int n = arr.Length; 
        int k = 3; 
  
        Console.WriteLine("Length = " + longSubarrWthSumDivByK(arr, n, k)); 
  
    } 
} 
  
// This code is contributed by Shrikant13 

Output:

Length = 4

Time Complexity: O(n).
Auxiliary Space: O(n^2).
Time complexity of this method can be improved by using an array of size equal to k for O(1) lookup since all elements would be less than k after using modulo operation on elements of input array.

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