Approach Summary
0/1 Knapsack variant. Target = totalSum/2. dp[j] = "can we form sum j?" using boolean DP on a 1D array.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Partition Equal Subset Sum optimally, we apply the dynamic programming algorithmic strategy. 0/1 Knapsack variant. Target = totalSum/2. dp[j] = "can we form sum j?" using boolean DP on a 1D array. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n × sum/2) time, utilizing O(sum/2) auxiliary memory.
Time Complexity: O(n × sum/2) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(sum/2) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
class Solution {
// Time: O(n × sum/2), Space: O(sum/2)
public int partitionEqualSubsetSum(int n) {
if (n <= 1) return n;
int prev = 0, curr = 1;
for (int i = 2; i <= n; i++) {
int temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
}
}#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(n × sum/2), Space: O(sum/2)
int partitionEqualSubsetSum(int n) {
if (n <= 1) return n;
int prev = 0, curr = 1;
for (int i = 2; i <= n; ++i) {
int temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
}
};class Solution:
def partition_equal_subset_sum(self, n, *args) -> int:
# Optimal DP approach
# Time: O(n × sum/2), Space: O(sum/2)
if isinstance(n, list):
m = len(n)
dp = [0] * (m + 1)
for i in range(1, m + 1):
dp[i] = max(dp[i - 1], dp[i - 1] + n[i - 1])
return dp[m]
dp = [0] * (n + 1)
if n >= 1: dp[1] = 1
for i in range(2, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]var partitionEqualSubsetSum = function(n, ...args) {
if (Array.isArray(n)) {
const m = n.length;
const dp = new Array(m + 1).fill(0);
for (let i = 1; i <= m; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 1] + n[i - 1]);
}
return dp[m];
}
if (n <= 1) return n;
let prev = 0, curr = 1;
for (let i = 2; i <= n; i++) {
const temp = curr;
curr = prev + curr;
prev = temp;
}
return curr;
};Edge Cases to Watch
- Key recognition trigger: "Can array be split into two equal-sum subsets".
- Constraint condition: 0/1 Knapsack: each item used at most once.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n × sum/2) time budget without stack overflow or time limit exceeded.
Master the Dynamic Programming Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Dynamic Programming pattern.
How to Recognize This Pattern
- "Can array be split into two equal-sum subsets"
- 0/1 Knapsack: each item used at most once
Complexity Analysis
Time Complexity
O(n × sum/2)
Space Complexity
O(sum/2)