1590. Make Sum Divisible by P
Description
Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.
Return the length of the smallest subarray that you need to remove, or -1 if it's impossible.
A subarray is defined as a contiguous block of elements in the array.
Example 1:
Input: nums = [3,1,4,2], p = 6 Output: 1 Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.
Example 2:
Input: nums = [6,3,5,2], p = 9 Output: 2 Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.
Example 3:
Input: nums = [1,2,3], p = 3 Output: 0 Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1091 <= p <= 109
Solutions
Solution 1: Prefix Sum + Hash Table
First, we calculate the sum of all elements in the array $\textit{nums}$ modulo $p$, denoted as $k$. If $k$ is $0$, it means the sum of all elements in the array $\textit{nums}$ is a multiple of $p$, so we directly return $0$.
If $k$ is not $0$, we need to find the shortest subarray such that removing this subarray makes the sum of the remaining elements modulo $p$ equal to $0$.
We can traverse the array $\textit{nums}$, maintaining the current prefix sum modulo $p$, denoted as $cur$. We use a hash table $last$ to record the last occurrence of each prefix sum modulo $p$.
If there exists a subarray ending at $\textit{nums}[i]$ such that removing this subarray makes the sum of the remaining elements modulo $p$ equal to $0$, we need to find a previous prefix sum modulo $p$ equal to $target$ at position $j$ such that $(target + k - cur) \bmod p = 0$. If found, we can remove the subarray $\textit{nums}[j+1,..i]$ to make the sum of the remaining elements modulo $p$ equal to $0$.
Therefore, if there exists a $target = (cur - k + p) \bmod p$, we can update the answer to $\min(ans, i - j)$. Then, we update $last[cur]$ to $i$. We continue traversing the array $\textit{nums}$ until the end to get the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
Python3
class Solution:
def minSubarray(self, nums: List[int], p: int) -> int:
k = sum(nums) % p
if k == 0:
return 0
last = {0: -1}
cur = 0
ans = len(nums)
for i, x in enumerate(nums):
cur = (cur + x) % p
target = (cur - k + p) % p
if target in last:
ans = min(ans, i - last[target])
last[cur] = i
return -1 if ans == len(nums) else ans
Java
class Solution {
public int minSubarray(int[] nums, int p) {
int k = 0;
for (int x : nums) {
k = (k + x) % p;
}
if (k == 0) {
return 0;
}
Map<Integer, Integer> last = new HashMap<>();
last.put(0, -1);
int n = nums.length;
int ans = n;
int cur = 0;
for (int i = 0; i < n; ++i) {
cur = (cur + nums[i]) % p;
int target = (cur - k + p) % p;
if (last.containsKey(target)) {
ans = Math.min(ans, i - last.get(target));
}
last.put(cur, i);
}
return ans == n ? -1 : ans;
}
}
C++
class Solution {
public:
int minSubarray(vector<int>& nums, int p) {
int k = 0;
for (int& x : nums) {
k = (k + x) % p;
}
if (k == 0) {
return 0;
}
unordered_map<int, int> last;
last[0] = -1;
int n = nums.size();
int ans = n;
int cur = 0;
for (int i = 0; i < n; ++i) {
cur = (cur + nums[i]) % p;
int target = (cur - k + p) % p;
if (last.count(target)) {
ans = min(ans, i - last[target]);
}
last[cur] = i;
}
return ans == n ? -1 : ans;
}
};
Go
func minSubarray(nums []int, p int) int {
k := 0
for _, x := range nums {
k = (k + x) % p
}
if k == 0 {
return 0
}
last := map[int]int{0: -1}
n := len(nums)
ans := n
cur := 0
for i, x := range nums {
cur = (cur + x) % p
target := (cur - k + p) % p
if j, ok := last[target]; ok {
ans = min(ans, i-j)
}
last[cur] = i
}
if ans == n {
return -1
}
return ans
}
TypeScript
function minSubarray(nums: number[], p: number): number {
let k = 0;
for (const x of nums) {
k = (k + x) % p;
}
if (k === 0) {
return 0;
}
const last = new Map<number, number>();
last.set(0, -1);
const n = nums.length;
let ans = n;
let cur = 0;
for (let i = 0; i < n; ++i) {
cur = (cur + nums[i]) % p;
const target = (cur - k + p) % p;
if (last.has(target)) {
const j = last.get(target)!;
ans = Math.min(ans, i - j);
}
last.set(cur, i);
}
return ans === n ? -1 : ans;
}
Rust
use std::collections::HashMap;
impl Solution {
pub fn min_subarray(nums: Vec<i32>, p: i32) -> i32 {
let mut k = 0;
for &x in &nums {
k = (k + x) % p;
}
if k == 0 {
return 0;
}
let mut last = HashMap::new();
last.insert(0, -1);
let n = nums.len();
let mut ans = n as i32;
let mut cur = 0;
for i in 0..n {
cur = (cur + nums[i]) % p;
let target = (cur - k + p) % p;
if let Some(&prev_idx) = last.get(&target) {
ans = ans.min(i as i32 - prev_idx);
}
last.insert(cur, i as i32);
}
if ans == n as i32 {
-1
} else {
ans
}
}
}
JavaScript
/**
* @param {number[]} nums
* @param {number} p
* @return {number}
*/
var minSubarray = function (nums, p) {
let k = 0;
for (const x of nums) {
k = (k + x) % p;
}
if (k === 0) {
return 0;
}
const last = new Map();
last.set(0, -1);
const n = nums.length;
let ans = n;
let cur = 0;
for (let i = 0; i < n; ++i) {
cur = (cur + nums[i]) % p;
const target = (cur - k + p) % p;
if (last.has(target)) {
const j = last.get(target);
ans = Math.min(ans, i - j);
}
last.set(cur, i);
}
return ans === n ? -1 : ans;
};