Array Partition - Problem
You're given an array of 2n integers and need to form n pairs to maximize a specific sum. Here's the challenge: for each pair (a, b), you only get to keep the smaller value min(a, b) toward your total score.
Goal: Group the integers into pairs such that the sum of all minimum values is as large as possible.
Example: With array [1, 4, 3, 2], you could pair as (1,4) and (3,2), giving you min(1,4) + min(3,2) = 1 + 2 = 3. But there's a better way!
Input: An integer array nums of length 2n
Output: The maximum possible sum of minimum values from optimal pairing
Input & Output
example_1.py — Basic Case
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Input:
[1,4,3,2]
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Output:
4
💡 Note:
Sort to get [1,2,3,4]. Pair as (1,2) and (3,4). Sum of minimums: min(1,2) + min(3,4) = 1 + 3 = 4.
example_2.py — All Same Values
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Input:
[6,2,6,5,1,2]
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Output:
9
💡 Note:
Sort to get [1,2,2,5,6,6]. Pair as (1,2), (2,5), (6,6). Sum of minimums: 1 + 2 + 6 = 9.
example_3.py — Minimum Case
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Input:
[1,2]
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Output:
1
💡 Note:
Only one pair possible: (1,2). Sum of minimums: min(1,2) = 1.
Constraints
- 1 ≤ n ≤ 104
- nums.length == 2 * n
- -104 ≤ nums[i] ≤ 104
- Array length is always even
Visualization
Tap to expand
Understanding the Visualization
1
Line Up Players
Sort all players by skill level from weakest to strongest
2
Form Adjacent Teams
Pair consecutive players: weakest with 2nd weakest, 3rd with 4th, etc.
3
Calculate Tournament Strength
Sum up each team's strength (their weaker player's skill)
Key Takeaway
🎯 Key Insight: Sort first, then pair consecutive elements. This greedy approach minimizes 'waste' by ensuring each element is paired with the closest possible partner in value.
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Explanation
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// Output will appear here after running code