Partition Equal Subset Sum
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Given an integer array, determine if it can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Example 1
- Input
- nums = [3,5,8,2,6]
- Output
- false
- Why
- Total sum is 24, target 12. No subset sums to 12.
Example 2
- Input
- nums = [4,7,1,5,3]
- Output
- true
- Why
- Total sum is 20, target 10. Subset [7,3] or [4,5,1] sums to 10.
Example 3
- Input
- nums = [1,1]
- Output
- true
- Why
- The array can be partitioned as [1] and [1].
Constraints
- 1 <= nums.length <= 200
- 1 <= nums[i] <= 100
- The sum of elements will not exceed 20000
How to think about it
Updated 2026-09-29Two equal subsets means each must sum to exactly half the total array sum. If the total sum is odd, equal partitioning is impossible from the start. Otherwise, the problem reduces to finding whether any subset sums to target = total / 2.
Approaches, worst first
Subsets branching
time O(2^n) · space O(n)
Recurse over each element with include and exclude decisions. Explores 2^n subsets, recalculating duplicate subset sums thousands of times.
Dynamic programming boolean arrayWrite this one
time O(n * sum) · space O(sum)
Compute target = sum / 2. Allocate boolean array dp of size target + 1 with dp[0] = true. For each num, sweep backwards from target down to num setting dp[j] = dp[j] || dp[j - num].
Where people lose marks · 3
- Forgetting the odd sum check. If total sum is odd, integer division truncates, leading to false positives for fractional targets.
- Iterating forward in the 1D DP table. Scanning from num up to target uses the same element multiple times to form its own sum.
- Running the search when the largest single number exceeds target. If max(nums) > target, no valid partition can exist.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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