Leetcode #2897: Apply Operations on Array to Maximize Sum of Squares
In this guide, we solve Leetcode #2897 Apply Operations on Array to Maximize Sum of Squares in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array nums and a positive integer k. You can do the following operation on the array any number of times: Choose any two distinct indices i and j and simultaneously update the values of nums[i] to (nums[i] AND nums[j]) and nums[j] to (nums[i] OR nums[j]).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Bit Manipulation, Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [2,6,5,8], k = 2
Output: 261
Explanation: We can do the following operations on the array:
- Choose i = 0 and j = 3, then change nums[0] to (2 AND 8) = 0 and nums[3] to (2 OR 8) = 10. The resulting array is nums = [0,6,5,10].
- Choose i = 2 and j = 3, then change nums[2] to (5 AND 10) = 0 and nums[3] to (5 OR 10) = 15. The resulting array is nums = [0,6,0,15].
We can choose the elements 15 and 6 from the final array. The sum of squares is 152 + 62 = 261.
It can be shown that this is the maximum value we can get.
Python Solution
class Solution:
def maxSum(self, nums: List[int], k: int) -> int:
mod = 10**9 + 7
cnt = [0] * 31
for x in nums:
for i in range(31):
if x >> i & 1:
cnt[i] += 1
ans = 0
for _ in range(k):
x = 0
for i in range(31):
if cnt[i]:
x |= 1 << i
cnt[i] -= 1
ans = (ans + x * x) % mod
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.