LeetCode #2601 — MEDIUM

Prime Subtraction Operation

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

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed integer array nums of length n.

You can perform the following operation as many times as you want:

  • Pick an index i that you haven’t picked before, and pick a prime p strictly less than nums[i], then subtract p from nums[i].

Return true if you can make nums a strictly increasing array using the above operation and false otherwise.

A strictly increasing array is an array whose each element is strictly greater than its preceding element.

Example 1:

Input: nums = [4,9,6,10]
Output: true
Explanation: In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0], so that nums becomes [1,9,6,10].
In the second operation: i = 1, p = 7, subtract 7 from nums[1], so nums becomes equal to [1,2,6,10].
After the second operation, nums is sorted in strictly increasing order, so the answer is true.

Example 2:

Input: nums = [6,8,11,12]
Output: true
Explanation: Initially nums is sorted in strictly increasing order, so we don't need to make any operations.

Example 3:

Input: nums = [5,8,3]
Output: false
Explanation: It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i] <= 1000
  • nums.length == n
Patterns Used
🔍Modified Binary Search→🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums of length n. You can perform the following operation as many times as you want: Pick an index i that you haven’t picked before, and pick a prime p strictly less than nums[i], then subtract p from nums[i]. Return true if you can make nums a strictly increasing array using the above operation and false otherwise. A strictly increasing array is an array whose each element is strictly greater than its preceding element.

Baseline thinking

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

Pattern signal: Array · Math · Binary Search · Greedy

Example 1

[4,9,6,10]

Example 2

[6,8,11,12]

Example 3

[5,8,3]
Step 02

Core Insight

What unlocks the optimal approach

  • Think about if we have many primes to subtract from nums[i]. Which prime is more optimal?
  • The most optimal prime to subtract from nums[i] is the one that makes nums[i] the smallest as possible and greater than nums[i-1].
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2601: Prime Subtraction Operation
class Solution {
    public boolean primeSubOperation(int[] nums) {
        List<Integer> p = new ArrayList<>();
        for (int i = 2; i <= 1000; ++i) {
            boolean ok = true;
            for (int j : p) {
                if (i % j == 0) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                p.add(i);
            }
        }
        int n = nums.length;
        for (int i = n - 2; i >= 0; --i) {
            if (nums[i] < nums[i + 1]) {
                continue;
            }
            int j = search(p, nums[i] - nums[i + 1]);
            if (j == p.size() || p.get(j) >= nums[i]) {
                return false;
            }
            nums[i] -= p.get(j);
        }
        return true;
    }

    private int search(List<Integer> nums, int x) {
        int l = 0, r = nums.size();
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums.get(mid) > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n log n)
Space
O(n)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.

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