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

• Difficulty Level : Medium
• Last Updated : 11 Jul, 2022

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:
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 sum 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++

 `// C++ Program to find the length of ` `// the longest subarray whose sum is ` `// not divisible by integer K ` ` `  `#include ` `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); ` `    ``int` `K = 3; ` ` `  `    ``cout << MaxSubarrayLength(arr, n, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

## Javascript

 ` `

Output:

`-1`

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

Related Topic: Subarrays, Subsequences, and Subsets in Array

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