Max Sum of Non-Adjacent Elements
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
The problem
Given an array of integers, find the maximum sum of non-adjacent elements. You cannot pick two elements that are next to each other in the array.
Example 1
- Input
- nums = [2,4,6,2,5]
- Output
- 13
- Why
- Pick elements 2, 6, and 5 for a sum of 13.
Example 2
- Input
- nums = [5,1,1,5]
- Output
- 10
- Why
- Pick the first 5 and the last 5 for a sum of 10.
Example 3
- Input
- nums = [1,2,3]
- Output
- 4
- Why
- Pick elements 1 and 3 for a sum of 4.
Constraints
- 1 <= nums.length <= 100
- 0 <= nums[i] <= 1000
How to think about it
Updated 2026-09-29Walking the array from left to right, maintaining whether each element is included or excluded reduces to a two-step choice: include the current element by adding it to the best sum excluding the previous element, or exclude it and take the maximum sum achievable up to the previous position.
Approaches, worst first
Exhaustive branching
time O(2^n) · space O(n)
Recurse over every element with a pick-or-skip decision at each step, returning the better of the two branches. It is a direct transcription of the problem and needs no insight to write, but the same suffix is re-solved once per distinct prefix that reaches it, so the work doubles with every extra element.
1D dynamic programming table
time O(n) · space O(n)
Create array dp of size n. dp[0] = nums[0], dp[1] = max(nums[0], nums[1]). For i from 2 to n - 1, dp[i] = max(dp[i-1], dp[i-2] + nums[i]). This tabulation avoids exponential recomputation but allocates an entire linear array when only the two preceding subproblem values are required.
Constant space state variablesWrite this one
time O(n) · space O(1)
Maintain two scalar accumulators: `include` (max sum including current element) and `exclude` (max sum excluding current element). At each step, new_include = exclude + num, and new_exclude = max(include, exclude).
Where people lose marks · 3
- Array with a single element: must immediately return nums[0] without accessing index 1.
- Overwriting exclude before calculating new_include. In-place variable updates must compute the new values simultaneously or use a temporary holder.
- Assuming alternating indices (all evens or all odds) gives the maximum. Skipping two elements in succession can yield a higher sum.
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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