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Length of smallest subarray to be removed to make sum of remaining elements divisible by K
• Difficulty Level : Hard
• Last Updated : 20 Nov, 2020

Given an array arr[] of integers and an integer K, the task is to find the length of the smallest subarray that needs to be removed such that the sum of remaining array elements is divisible by K. Removal of the entire array is not allowed. If it is impossible, then print “-1”.

Examples:

Input: arr[] = {3, 1, 4, 2}, K = 6
Output: 1
Explanation: Sum of array elements = 10, which is not divisible by 6. After removing the subarray {4}, sum of the remaining elements is 6. Therefore, the length of the removed subarray is 1.

Input: arr[] = {3, 6, 7, 1}, K = 9
Output: 2
Explanation: Sum of array elements = 17, which is not divisible by 9. After removing the subarray {7, 1} and the, sum of the remaining elements is 9. Therefore, the length of the removed subarray is 2.

Naive Approach: The simplest approach is to generate all possible subarray from the given array arr[] excluding the subarray of length N. Now, find the minimum length of subarray such that the difference between the sum of all the elements of the array and the sum of the elements in that subarray is divisible by K. If no such subarray exists, then print “-1”

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the below observation:

((total_sum – subarray_sum) % K + subarray_sum % K) must be equal to total_sum % K.
But, (total_sum – subarray_sum) % K == 0 should be true.

Therefore, total_sum % K == subarray_sum % K, so both subarray_sum and total_sum should leave the same remainder when divided by K. Hence, the task is to find the length of the smallest subarray whose sum of elements will leave a remainder of (total_sum % K).

Follow the steps below to solve this problem:

• Initialize variable res as INT_MAX to store the minimum length of the subarray to be removed.
• Calculate total_sum and the remainder which it leaves when divided by K.
• Create an auxiliary array modArr[] to storing the remainder of each arr[i] when it is divided by K as:

modArr[i] = (arr[i] + K) % K.
where,
K has been added while calculating the remainder to handle the case of negative integers.

• Traverse the given array and maintain an unordered_map to stores the recent position of the remainder encountered and keep track of the minimum required subarray having the remainder same as the target_remainder.
• If there exists any key in the map which is equal to (curr_remainder – target_remainder + K) % K, then store that subarray length in variable res as the minimum of res and current length found.
• After the above, if res is unchanged the print “-1” Otherwise print the value of res.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find the length of the``// smallest subarray to be removed such``// that sum of elements is divisible by K``void` `removeSmallestSubarray(``int` `arr[],``                            ``int` `n, ``int` `k)``{``    ``// Stores the remainder of each``    ``// arr[i] when divided by K``    ``int` `mod_arr[n];` `    ``// Stores total sum of elements``    ``int` `total_sum = 0;` `    ``// K has been added to each arr[i]``    ``// to handle -ve integers``    ``for` `(``int` `i = 0; i < n; i++) {``        ``mod_arr[i] = (arr[i] + k) % k;` `        ``// Update the total sum``        ``total_sum += arr[i];``    ``}` `    ``// Remainder when total_sum``    ``// is divided by K``    ``int` `target_remainder``        ``= total_sum % k;` `    ``// If given array is already``    ``// divisible by K``    ``if` `(target_remainder == 0) {``        ``cout << ``"0"``;``        ``return``;``    ``}` `    ``// Stores curr_remainder and the``    ``// most recent index at which``    ``// curr_remainder has occured``    ``unordered_map<``int``, ``int``> map1;``    ``map1[0] = -1;` `    ``int` `curr_remainder = 0;` `    ``// Stores required answer``    ``int` `res = INT_MAX;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Add current element to``        ``// curr_sum and take mod``        ``curr_remainder = (curr_remainder``                          ``+ arr[i] + k)``                         ``% k;` `        ``// Update current remainder index``        ``map1[curr_remainder] = i;` `        ``int` `mod``            ``= (curr_remainder``               ``- target_remainder``               ``+ k)``              ``% k;` `        ``// If mod already exists in map``        ``// the subarray exists``        ``if` `(map1.find(mod) != map1.end())``            ``res = min(res, i - map1[mod]);``    ``}` `    ``// If not possible``    ``if` `(res == INT_MAX || res == n) {``        ``res = -1;``    ``}` `    ``// Print the result``    ``cout << res;``}` `// Driver Code``int` `main()``{``    ``// Given array arr[]``    ``int` `arr[] = { 3, 1, 4, 2 };` `    ``// Size of array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Given K``    ``int` `K = 6;` `    ``// Function Call``    ``removeSmallestSubarray(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the``// above approach``import` `java.util.*;``class` `GFG{` `// Function to find the length of the``// smallest subarray to be removed such``// that sum of elements is divisible by K``static` `void` `removeSmallestSubarray(``int` `arr[],``                                   ``int` `n, ``int` `k)``{``  ``// Stores the remainder of each``  ``// arr[i] when divided by K``  ``int` `[]mod_arr = ``new` `int``[n];` `  ``// Stores total sum of``  ``// elements``  ``int` `total_sum = ``0``;` `  ``// K has been added to each``  ``// arr[i] to handle -ve integers``  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``mod_arr[i] = (arr[i] +``                  ``k) % k;` `    ``// Update the total sum``    ``total_sum += arr[i];``  ``}` `  ``// Remainder when total_sum``  ``// is divided by K``  ``int` `target_remainder =``      ``total_sum % k;` `  ``// If given array is already``  ``// divisible by K``  ``if` `(target_remainder == ``0``)``  ``{``    ``System.out.print(``"0"``);``    ``return``;``  ``}` `  ``// Stores curr_remainder and the``  ``// most recent index at which``  ``// curr_remainder has occured``  ``HashMap map1 =``          ``new` `HashMap<>();``  ``map1.put(``0``, -``1``);` `  ``int` `curr_remainder = ``0``;` `  ``// Stores required answer``  ``int` `res = Integer.MAX_VALUE;` `  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``// Add current element to``    ``// curr_sum and take mod``    ``curr_remainder = (curr_remainder +``                      ``arr[i] + k) % k;` `    ``// Update current remainder``    ``// index``    ``map1.put(curr_remainder, i);` `    ``int` `mod = (curr_remainder -``               ``target_remainder +``               ``k) % k;` `    ``// If mod already exists in``    ``// map the subarray exists``    ``if` `(map1.containsKey(mod))``      ``res = Math.min(res, i -``                     ``map1.get(mod));``  ``}` `  ``// If not possible``  ``if` `(res == Integer.MAX_VALUE ||``      ``res == n)``  ``{``    ``res = -``1``;``  ``}` `  ``// Print the result``  ``System.out.print(res);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``// Given array arr[]``  ``int` `arr[] = {``3``, ``1``, ``4``, ``2``};` `  ``// Size of array``  ``int` `N = arr.length;` `  ``// Given K``  ``int` `K = ``6``;` `  ``// Function Call``  ``removeSmallestSubarray(arr, N, K);``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Python3 program for the above approach``import` `sys` `# Function to find the length of the``# smallest subarray to be removed such``# that sum of elements is divisible by K``def` `removeSmallestSubarray(arr, n, k):``    ` `    ``# Stores the remainder of each``    ``# arr[i] when divided by K``    ``mod_arr ``=` `[``0``] ``*` `n`` ` `    ``# Stores total sum of elements``    ``total_sum ``=` `0`` ` `    ``# K has been added to each arr[i]``    ``# to handle -ve integers``    ``for` `i ``in` `range``(n) :``        ``mod_arr[i] ``=` `(arr[i] ``+` `k) ``%` `k`` ` `        ``# Update the total sum``        ``total_sum ``+``=` `arr[i]``        ` `    ``# Remainder when total_sum``    ``# is divided by K``    ``target_remainder ``=` `total_sum ``%` `k`` ` `    ``# If given array is already``    ``# divisible by K``    ``if` `(target_remainder ``=``=` `0``):``        ``print``(``"0"``)``        ``return``    ` `    ``# Stores curr_remainder and the``    ``# most recent index at which``    ``# curr_remainder has occured``    ``map1 ``=` `{}``    ``map1[``0``] ``=` `-``1`` ` `    ``curr_remainder ``=` `0`` ` `    ``# Stores required answer``    ``res ``=` `sys.maxsize`` ` `    ``for` `i ``in` `range``(n):``        ` `        ``# Add current element to``        ``# curr_sum and take mod``        ``curr_remainder ``=` `(curr_remainder ``+``                             ``arr[i] ``+` `k) ``%` `k`` ` `        ``# Update current remainder index``        ``map1[curr_remainder] ``=` `i`` ` `        ``mod ``=` `(curr_remainder ``-``             ``target_remainder ``+` `k) ``%` `k`` ` `        ``# If mod already exists in map``        ``# the subarray exists``        ``if` `(mod ``in` `map1.keys()):``            ``res ``=` `min``(res, i ``-` `map1[mod])``    ` `    ``# If not possible``    ``if` `(res ``=``=` `sys.maxsize ``or` `res ``=``=` `n):``        ``res ``=` `-``1``    ` `    ``# Print the result``    ``print``(res)` `# Driver Code` `# Given array arr[]``arr ``=` `[ ``3``, ``1``, ``4``, ``2` `]`` ` `# Size of array``N ``=` `len``(arr)`` ` `# Given K``K ``=` `6`` ` `# Function Call``removeSmallestSubarray(arr, N, K)` `# This code is contributed by susmitakundugoaldanga`

## C#

 `// C# program for the``// above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the length of the``// smallest subarray to be removed such``// that sum of elements is divisible by K``static` `void` `removeSmallestSubarray(``int` `[]arr,``                                   ``int` `n, ``int` `k)``{``  ` `  ``// Stores the remainder of each``  ``// arr[i] when divided by K``  ``int` `[]mod_arr = ``new` `int``[n];` `  ``// Stores total sum of``  ``// elements``  ``int` `total_sum = 0;` `  ``// K has been added to each``  ``// arr[i] to handle -ve integers``  ``for``(``int` `i = 0; i < n; i++)``  ``{``    ``mod_arr[i] = (arr[i] + k) % k;` `    ``// Update the total sum``    ``total_sum += arr[i];``  ``}` `  ``// Remainder when total_sum``  ``// is divided by K``  ``int` `target_remainder = total_sum % k;` `  ``// If given array is already``  ``// divisible by K``  ``if` `(target_remainder == 0)``  ``{``    ``Console.Write(``"0"``);``    ``return``;``  ``}` `  ``// Stores curr_remainder and the``  ``// most recent index at which``  ``// curr_remainder has occured``  ``Dictionary<``int``,``             ``int``> map1 = ``new` `Dictionary<``int``,``                                        ``int``>();``  ` `  ``map1.Add(0, -1);` `  ``int` `curr_remainder = 0;` `  ``// Stores required answer``  ``int` `res = ``int``.MaxValue;` `  ``for``(``int` `i = 0; i < n; i++)``  ``{``    ` `    ``// Add current element to``    ``// curr_sum and take mod``    ``curr_remainder = (curr_remainder +``                      ``arr[i] + k) % k;` `    ``// Update current remainder``    ``// index``    ``map1[curr_remainder] = i;` `    ``int` `mod = (curr_remainder -``               ``target_remainder +``               ``k) % k;` `    ``// If mod already exists in``    ``// map the subarray exists``    ``if` `(map1.ContainsKey(mod))``      ``res = Math.Min(res, i -``                     ``map1[mod]);``  ``}` `  ``// If not possible``  ``if` `(res == ``int``.MaxValue ||``      ``res == n)``  ``{``    ``res = -1;``  ``}``  ` `  ``// Print the result``  ``Console.Write(res);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ` `  ``// Given array []arr``  ``int` `[]arr = { 3, 1, 4, 2 };` `  ``// Size of array``  ``int` `N = arr.Length;` `  ``// Given K``  ``int` `K = 6;` `  ``// Function Call``  ``removeSmallestSubarray(arr, N, K);``}``}` `// This code is contributed by 29AjayKumar`
Output:
```1

```

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

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