LeetCode #698 — MEDIUM

Partition to K Equal Sum Subsets

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

The Problem

Problem Statement

Given an integer array nums and an integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal.

Example 1:

Input: nums = [4,3,2,3,5,2,1], k = 4
Output: true
Explanation: It is possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.

Example 2:

Input: nums = [1,2,3,4], k = 3
Output: false

Constraints:

1 <= k <= nums.length <= 16
1 <= nums[i] <= 10⁴
The frequency of each element is in the range [1, 4].

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums and an integer k, return true if it is possible to divide this array into k non-empty subsets whose sums are all equal.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[4,3,2,3,5,2,1]
4

Example 2

[1,2,3,4]
3

Core Insight

What unlocks the optimal approach

We can figure out what target each subset must sum to. Then, let's recursively search, where at each call to our function, we choose which of k subsets the next value will join.

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 #698: Partition to K Equal Sum Subsets
class Solution {
    private int[] nums;
    private int[] cur;
    private int s;

    public boolean canPartitionKSubsets(int[] nums, int k) {
        for (int v : nums) {
            s += v;
        }
        if (s % k != 0) {
            return false;
        }
        s /= k;
        cur = new int[k];
        Arrays.sort(nums);
        this.nums = nums;
        return dfs(nums.length - 1);
    }

    private boolean dfs(int i) {
        if (i < 0) {
            return true;
        }
        for (int j = 0; j < cur.length; ++j) {
            if (j > 0 && cur[j] == cur[j - 1]) {
                continue;
            }
            cur[j] += nums[i];
            if (cur[j] <= s && dfs(i - 1)) {
                return true;
            }
            cur[j] -= nums[i];
        }
        return false;
    }
}

// Accepted solution for LeetCode #698: Partition to K Equal Sum Subsets
func canPartitionKSubsets(nums []int, k int) bool {
	s := 0
	for _, v := range nums {
		s += v
	}
	if s%k != 0 {
		return false
	}
	s /= k
	cur := make([]int, k)
	n := len(nums)

	var dfs func(int) bool
	dfs = func(i int) bool {
		if i == n {
			return true
		}
		for j := 0; j < k; j++ {
			if j > 0 && cur[j] == cur[j-1] {
				continue
			}
			cur[j] += nums[i]
			if cur[j] <= s && dfs(i+1) {
				return true
			}
			cur[j] -= nums[i]
		}
		return false
	}

	sort.Sort(sort.Reverse(sort.IntSlice(nums)))
	return dfs(0)
}

# Accepted solution for LeetCode #698: Partition to K Equal Sum Subsets
class Solution:
    def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
        def dfs(i: int) -> bool:
            if i == len(nums):
                return True
            for j in range(k):
                if j and cur[j] == cur[j - 1]:
                    continue
                cur[j] += nums[i]
                if cur[j] <= s and dfs(i + 1):
                    return True
                cur[j] -= nums[i]
            return False

        s, mod = divmod(sum(nums), k)
        if mod:
            return False
        cur = [0] * k
        nums.sort(reverse=True)
        return dfs(0)

// Accepted solution for LeetCode #698: Partition to K Equal Sum Subsets
struct Solution;

impl Solution {
    fn can_partition_k_subsets(nums: Vec<i32>, k: i32) -> bool {
        let n = nums.len();
        let sum: i32 = nums.iter().sum();
        if sum % k != 0 {
            return false;
        }
        let mut visited: Vec<bool> = vec![false; n];
        Self::search(0, 0, k as usize, &mut visited, &nums, n, sum / k)
    }

    fn search(
        start: usize,
        sum: i32,
        k: usize,
        visited: &mut Vec<bool>,
        nums: &[i32],
        n: usize,
        target: i32,
    ) -> bool {
        if k == 0 {
            return true;
        }
        for i in start..n {
            if visited[i] {
                continue;
            }
            visited[i] = true;
            if sum + nums[i] < target
                && Self::search(i + 1, sum + nums[i], k, visited, nums, n, target)
            {
                return true;
            }
            if sum + nums[i] == target && Self::search(0, 0, k - 1, visited, nums, n, target) {
                return true;
            }
            visited[i] = false;
        }
        false
    }
}

#[test]
fn test() {
    // let nums = vec![4, 3, 2, 3, 5, 2, 1];
    // let k = 4;
    // let res = true;
    // assert_eq!(Solution::can_partition_k_subsets(nums, k), res);
    let nums = vec![2, 2, 2, 2, 3, 4, 5];
    let k = 4;
    let res = false;
    assert_eq!(Solution::can_partition_k_subsets(nums, k), res);
}

// Accepted solution for LeetCode #698: Partition to K Equal Sum Subsets
function canPartitionKSubsets(nums: number[], k: number): boolean {
    const dfs = (i: number): boolean => {
        if (i === nums.length) {
            return true;
        }
        for (let j = 0; j < k; j++) {
            if (j > 0 && cur[j] === cur[j - 1]) {
                continue;
            }
            cur[j] += nums[i];
            if (cur[j] <= s && dfs(i + 1)) {
                return true;
            }
            cur[j] -= nums[i];
        }
        return false;
    };

    let s = nums.reduce((a, b) => a + b, 0);
    const mod = s % k;
    if (mod !== 0) {
        return false;
    }
    s = Math.floor(s / k);
    const cur = Array(k).fill(0);
    nums.sort((a, b) => b - a);
    return dfs(0);
}

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 × 2^n)

Space

O(2^n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.