LeetCode #1590 — MEDIUM

Make Sum Divisible by P

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 <= 10⁵
1 <= nums[i] <= 10⁹
1 <= p <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[3,1,4,2]
6

Example 2

[6,3,5,2]
9

Example 3

[1,2,3]
3

Core Insight

What unlocks the optimal approach

Use prefix sums to calculate the subarray sums.
Suppose you know the remainder for the sum of the entire array. How does removing a subarray affect that remainder? What remainder does the subarray need to have in order to make the rest of the array sum up to be divisible by k?
Use a map to keep track of the rightmost index for every prefix sum % p.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1590: Make Sum Divisible by P
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;
    }
}

// Accepted solution for LeetCode #1590: Make Sum Divisible by P
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
}

# Accepted solution for LeetCode #1590: Make Sum Divisible by P
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

// Accepted solution for LeetCode #1590: Make Sum Divisible by P
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
        }
    }
}

// Accepted solution for LeetCode #1590: Make Sum Divisible by P
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.