LeetCode #3566 — MEDIUM

Partition Array into Two Equal Product Subsets

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

The Problem

Problem Statement

You are given an integer array nums containing distinct positive integers and an integer target.

Determine if you can partition nums into two non-empty disjoint subsets, with each element belonging to exactly one subset, such that the product of the elements in each subset is equal to target.

Return true if such a partition exists and false otherwise.

A subset of an array is a selection of elements of the array.

Example 1:

Input: nums = [3,1,6,8,4], target = 24

Output: true

Explanation: The subsets [3, 8] and [1, 6, 4] each have a product of 24. Hence, the output is true.

Example 2:

Input: nums = [2,5,3,7], target = 15

Output: false

Explanation: There is no way to partition nums into two non-empty disjoint subsets such that both subsets have a product of 15. Hence, the output is false.

Constraints:

3 <= nums.length <= 12
1 <= target <= 10¹⁵
1 <= nums[i] <= 100
All elements of nums are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums containing distinct positive integers and an integer target. Determine if you can partition nums into two non-empty disjoint subsets, with each element belonging to exactly one subset, such that the product of the elements in each subset is equal to target. Return true if such a partition exists and false otherwise. A subset of an array is a selection of elements of the array.

Baseline thinking

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

Pattern signal: Array · Bit Manipulation

Example 1

[3,1,6,8,4]
24

Example 2

[2,5,3,7]
15

Step 02

Core Insight

What unlocks the optimal approach

Try iterating over all subsets
Use bitmasks

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 #3566: Partition Array into Two Equal Product Subsets
class Solution {
    public boolean checkEqualPartitions(int[] nums, long target) {
        int n = nums.length;
        for (int i = 0; i < 1 << n; ++i) {
            long x = 1, y = 1;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    x *= nums[j];
                } else {
                    y *= nums[j];
                }
            }
            if (x == target && y == target) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #3566: Partition Array into Two Equal Product Subsets
func checkEqualPartitions(nums []int, target int64) bool {
	n := len(nums)
	for i := 0; i < 1<<n; i++ {
		x, y := int64(1), int64(1)
		for j, v := range nums {
			if i>>j&1 == 1 {
				x *= int64(v)
			} else {
				y *= int64(v)
			}
			if x > target || y > target {
				break
			}
		}
		if x == target && y == target {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #3566: Partition Array into Two Equal Product Subsets
class Solution:
    def checkEqualPartitions(self, nums: List[int], target: int) -> bool:
        n = len(nums)
        for i in range(1 << n):
            x = y = 1
            for j in range(n):
                if i >> j & 1:
                    x *= nums[j]
                else:
                    y *= nums[j]
            if x == target and y == target:
                return True
        return False

// Accepted solution for LeetCode #3566: Partition Array into Two Equal Product Subsets
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3566: Partition Array into Two Equal Product Subsets
// class Solution {
//     public boolean checkEqualPartitions(int[] nums, long target) {
//         int n = nums.length;
//         for (int i = 0; i < 1 << n; ++i) {
//             long x = 1, y = 1;
//             for (int j = 0; j < n; ++j) {
//                 if ((i >> j & 1) == 1) {
//                     x *= nums[j];
//                 } else {
//                     y *= nums[j];
//                 }
//             }
//             if (x == target && y == target) {
//                 return true;
//             }
//         }
//         return false;
//     }
// }

// Accepted solution for LeetCode #3566: Partition Array into Two Equal Product Subsets
function checkEqualPartitions(nums: number[], target: number): boolean {
    const n = nums.length;
    for (let i = 0; i < 1 << n; ++i) {
        let [x, y] = [1, 1];
        for (let j = 0; j < n; ++j) {
            if (((i >> j) & 1) === 1) {
                x *= nums[j];
            } else {
                y *= nums[j];
            }
            if (x > target || y > target) {
                break;
            }
        }
        if (x === target && y === target) {
            return true;
        }
    }
    return false;
}

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(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.