Delete Sublist to Make Sum Divisible By K - Amazon Top Interview Questions


Problem Statement :


You are given a list of positive integers nums and a positive integer k. Return the length of the shortest sublist (can be empty sublist ) you can delete such that the resulting list's sum is divisible by k. You cannot delete the entire list. If it's not possible, return -1.

Constraints

1 ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 8, 6, 4, 5]

k = 7

Output

2

Explanation

We can remove the sublist [6, 4] to get [1, 8, 5] which sums to 14 and is divisible by 7.


Solution :



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                        Solution in C++ :

int solve(vector<int>& nums, int k) {
    if (nums.empty()) return -1;

    int target = 0;
    for (int num : nums) target = (target + num) % k;
    if (target == 0) return 0;

    unordered_map<int, int> memo;
    memo[0] = 0;
    int pos = 0, now = 0, ans = nums.size();
    for (int num : nums) {
        pos++;
        now = (now + num) % k;
        if (memo.count((now + k - target) % k)) ans = min(ans, pos - memo[(now + k - target) % k]);
        memo[now] = pos;
    }
    return ans == nums.size() ? -1 : ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int k) {
        //(totalSUm - delete) % k == 0;
        // totalSUm % k = delet%k
        // let totalSum % k = totalMod
        // totalMod % k = totalMod;
        // delet%k = prefix[j] % k - prefix[i] %k = totalMod %k;
        // pre[i] % k = (pre[j] - totalMod) % k;
        int total = 0;
        for (int n : nums) {
            total += n;
        }
        int totalMod = Math.floorMod(total, k);

        Map<Integer, Integer> map = new HashMap();
        map.put(0, -1);
        int prefix = 0;
        int res = nums.length;
        for (int i = 0; i < nums.length; i++) {
            prefix += nums[i];
            int curMod = Math.floorMod(prefix, k);
            map.put(curMod, i);
            int afterMod = Math.floorMod(prefix - totalMod, k);
            // System.out.println(afterMod + "total" + totalMod +"pre" + prefix);
            if (map.containsKey(afterMod)) {
                int len = i - map.get(afterMod);
                System.out.println(i);
                res = len < res ? len : res;
            }
        }
        return res == nums.length ? -1 : res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, k):
        tar_remainder = (sum(nums) + k) % k
        if tar_remainder == 0:
            return 0
        n, presum = len(nums), 0
        hashmap = {0: -1}
        res = n
        for i in range(n):
            presum += nums[i]
            modulus = (presum + k) % k
            hashmap[modulus] = i
            if (modulus - tar_remainder + k) % k in hashmap:
                res = min(res, i - hashmap[(modulus - tar_remainder + k) % k])
        return res if res != n else -1


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