Maximize Sum of Device Ratings - Video Solutions
Leetcode 3961 Weekly Contest 506 Q3 | Maximize Sum of Device Ratings | Easy Solution🔥
Maximize Sum of Device Ratings - Video Solution
Watch 7 video solutions for Maximize Sum of Device Ratings, a medium level problem involving Array, Greedy, Sorting. This walkthrough by CodeSprint has 542 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
You are given a 2D integer array units of size m × n where units[i][j] represents the capacity of the jth unit in the ith device. Each device contains exactly n units.
The rating of a device is the minimum capacity among all its units.
You may perform the following operation any number of times (including zero):
- Choose a device
ithat has not been used as a source before. - Remove exactly one unit from device
iand add it to any different device. - Then mark device
ias used, so it cannot be chosen again as a source.
Return the maximum possible sum of the ratings of all devices after any number of such operations.
Note:
- Devices can receive units from multiple devices, regardless of whether they have been selected.
- The rating of an empty device is 0.
Example 1:
Input: units = [[1,3],[2,2]]
Output: 4
Explanation:
- ​​​​​​​​​​​​​​Select device
i = 0and transferunits[0][0] = 1to devicei = 1. - After the transfer, the ratings are:
- Device
0 = [3]:rating[0] = 3 - Device
1 = [2, 2, 1]:rating[1] = 1
- Device
- Thus, the sum of ratings is
3 + 1 = 4.
Example 2:
Input: units = [[1,2,3],[4,5,6]]
Output: 6
Explanation:
- Select device
i = 1and transferunits[1][0] = 4to devicei = 0. - After the transfer, the ratings are:
- Device
0 = [1, 2, 3, 4]:rating[0] = 1 - Device
1 = [5, 6]:rating[1] = 5
- Device
- Thus, the sum of ratings is
1 + 5 = 6.
Example 3:
Input: units = [[5,5,5],[1,1,1]]
Output: 6
Explanation:
- No transfers increase the sum of ratings. Thus, the sum of ratings is
5 + 1 = 6.
Constraints:
1 <= m == units.length <= 1051 <= n == units[i].length <= 105m * n <= 2 * 1051 <= units[i][j] <= 105
Approach Overview
Problem Overview: You are given an array representing device ratings. Activating certain devices may restrict activating adjacent ones, so the goal is to choose a subset of devices that maximizes the total rating while respecting the constraint.
Approach 1: Brute Force Recursion (Exponential Time, O(2^n) time, O(n) space)
Try every possible subset of devices. At each index, you either activate the current device or skip it. If you activate it, the next device cannot be chosen, so the recursion jumps two indices forward. If you skip it, continue with the next index. This approach explores the full decision tree and guarantees the correct answer, but the repeated recalculation of overlapping subproblems causes exponential runtime.
Approach 2: Dynamic Programming with Memoization (O(n) time, O(n) space)
The recursive solution repeatedly solves the same states. Store results in a memo array where dp[i] represents the maximum rating achievable starting from index i. For each position, compute max(rating[i] + dp[i+2], dp[i+1]). Each state is evaluated once, reducing complexity to linear time. This method clearly exposes the overlapping subproblem structure typical in dynamic programming problems.
Approach 3: Bottom-Up DP (O(n) time, O(n) space)
Instead of recursion, build the solution iteratively. Define dp[i] as the best sum achievable considering the first i devices. The transition becomes dp[i] = max(dp[i-1], dp[i-2] + rating[i]). Iterate through the array once while maintaining this relationship. This version avoids recursion overhead and is easier to reason about during interviews. The approach relies on recognizing optimal substructure within the array.
Approach 4: Space Optimized DP (O(n) time, O(1) space)
The DP transition only depends on the previous two states. Instead of maintaining the entire DP array, track two variables representing dp[i-1] and dp[i-2]. For each device rating, compute the new best value and shift the variables forward. This keeps memory usage constant while preserving the same linear runtime. The logic mirrors classic patterns seen in many dynamic programming optimization problems.
Recommended for interviews: Start by describing the brute force choice of picking or skipping a device to show understanding of the problem structure. Then transition to dynamic programming, which interviewers typically expect. The space‑optimized DP solution demonstrates strong problem‑solving ability because it recognizes that only two previous states are required.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Recursion | O(2^n) | O(n) | Useful for understanding the decision tree and problem structure |
| DP with Memoization | O(n) | O(n) | General solution when using recursion with caching |
| Bottom-Up Dynamic Programming | O(n) | O(n) | Preferred iterative DP approach in interviews |
| Space Optimized DP | O(n) | O(1) | When memory efficiency matters and only previous states are required |