Partition Array for Maximum Sum

Partition Array for Maximum Sum

What is this problem about?

The Partition Array for Maximum Sum problem asks you to partition an array into subarrays of length at most k, replacing each subarray with its maximum value repeated for the subarray's length. Maximize the total sum. This coding problem uses DP where each position tries all valid partition end points. The array and dynamic programming interview pattern is the core.

Why is this asked in interviews?

Apple, Microsoft, Meta, Amazon, Google, and Bloomberg ask this because it tests partition DP where the "cost" of a partition depends on the maximum element. The ability to define the DP transition for a max-based partition shows DP mastery.

Algorithmic pattern used

Partition DP. dp[i] = maximum sum for the first i elements. For each i, try last partition of length 1..k: dp[i] = max(dp[i-len] + max(arr[i-len..i-1]) * len) for len=1..min(k,i). Answer = dp[n].

Example explanation

arr=[1,15,7,9,2,5,10], k=3. dp[1]=1. dp[2]=max(dp[1]+15, dp[0]+152)=max(16,30)=30. dp[3]=max(dp[2]+7, dp[1]+152, dp[0]+15*3)=max(37,31,45)=45. Continue computing dp. dp[7] = 84.

Common mistakes candidates make

• Not tracking the running maximum while extending the partition backward.
• Off-by-one in the partition length loop.
• Overwriting dp values incorrectly.
• Not initializing dp[0]=0 (empty array contributes 0).

Interview preparation tip

"Partition with max-based cost" DPs are a common pattern. The key: for each ending position i, try all starting points j = i-1, i-2, ..., i-k. Track the running max as you extend the partition left. The DP value = max(dp[j] + max(j..i-1) * (i-j)) over all valid j. Practice similar partition DP problems: "paint houses," "split array largest sum," "burst balloons." Each has a different "cost" function but the same DP structure.