What is the brute force approach and complexity for Make Sum Divisible by P?

Try every subarray; keep the shortest whose sum % p == need with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Make Sum Divisible by P?

Look up remainder (cur − need) mod p; store the latest index of each remainder with optimal time complexity O(N) and space complexity O(min(N, P)).

Make Sum Divisible by P

Q: How do you identify the pattern for Make Sum Divisible by P?

Look for key signals in the prompt: "remove a contiguous subarray" indicates The removed piece is a range → difference of two prefix sums.; "rest divisible by p" indicates The piece must have the same remainder as the total: `need = total % p`. Key the map by prefix remainder.; "smallest length" indicates We want the closest matching start → store the **last** index of each remainder.; "not the whole array / return −1" indicates A match of length n does not count; check it at the end..

Medium·LeetCode #1590 Pro OnlyTricky / Edge Case

LeetCode

Problem Statement

You are given an array nums of positive integers and a number p.

You may delete one contiguous subarray (possibly empty) so that the sum of what is left is divisible by p. You are not allowed to delete the whole array.

Return the length of the shortest subarray you need to delete, or -1 if it cannot be done.

Input

An integer array nums (all values ≥ 1) and an integer p (p ≥ 1).

Output

The smallest length of a subarray to remove (0 if the sum is already divisible), or −1 if only removing everything would work.

Example 1

Input: nums = [3, 1, 4, 2], p = 6

Output: 1

The total is 10. Remove [4] and the rest sums to 6.

Example 2

Input: nums = [6, 3, 5, 2], p = 9

Output: 2

The total is 16. No single element works, but removing [5, 2] leaves [6, 3] with sum 9.

Example 3

Input: nums = [1, 2, 3], p = 3

Output: 0

The total 6 is already divisible by 3, so remove nothing.

Example 4

Input: nums = [1, 2, 3], p = 7

Output: -1

The total is 6. Only removing the whole array would leave a sum divisible by 7, and that is not allowed.

Constraints

• 1 ≤ nums.length ≤ 10^5
• 1 ≤ nums[i] ≤ 10^9
• 1 ≤ p ≤ 10^9
• The total can reach 10^14 — use a 64-bit type or keep everything mod p.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“remove a contiguous subarray”

→ The removed piece is a range → difference of two prefix sums.

“rest divisible by p”

→ The piece must have the same remainder as the total: `need = total % p`. Key the map by prefix remainder.

“smallest length”

→ We want the closest matching start → store the **last** index of each remainder.

“not the whole array / return −1”

→ A match of length n does not count; check it at the end.

Approach Complexity Ladder

Brute Force — try every piece to removeO(N²)O(1)

Try every subarray; keep the shortest whose sum % p == need

Optimal — last index of each prefix remainderO(N)O(min(N, P))

Look up remainder (cur − need) mod p; store the latest index of each remainder

Brute Force Intuition

Compute need = total % p. For every start, grow the end and keep the piece's remainder. Whenever it equals need, the rest is divisible by p, so record the length. Ignore the whole array.

Recognition rule for this sub-pattern:

Sum divisible by K ⇔ two prefixes have the same remainder. Store remainders, not sums. Fix negative remainders.

Remember this · Prefix Sum

Trigger

Range sums, "subarray sum equals K" (especially with negatives), equal counts, divisible by K.

Think

If I knew the total up to every index, would the answer become a difference of two totals?

Key question

Which earlier prefix value am I looking for when I stand at index i?

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