LeetCode #2897 — HARD

Apply Operations on Array to Maximize Sum of Squares

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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]). Here, OR denotes the bitwise OR operation, and AND denotes the bitwise AND operation.

You have to choose k elements from the final array and calculate the sum of their squares.

Return the maximum sum of squares you can achieve.

Since the answer can be very large, return it modulo 10⁹ + 7.

Example 1:

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.

Example 2:

Input: nums = [4,5,4,7], k = 3
Output: 90
Explanation: We do not need to apply any operations.
We can choose the elements 7, 5, and 4 with a sum of squares: 7² + 5² + 4² = 90.
It can be shown that this is the maximum value we can get.

Constraints:

1 <= k <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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]). Here, OR denotes the bitwise OR operation, and AND denotes the bitwise AND operation. You have to choose k elements from the final array and calculate the sum of their squares. Return the maximum sum of squares you can achieve. Since the answer can be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy · Bit Manipulation

Example 1

[2,6,5,8]
2

Example 2

[4,5,4,7]
3

Core Insight

What unlocks the optimal approach

The operation described only transfers some bits from one element to another in their binary representation.
To have a maximum sum of squares, it is optimal to greedily make each number as big as possible.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
class Solution {
    public int maxSum(List<Integer> nums, int k) {
        final int mod = (int) 1e9 + 7;
        int[] cnt = new int[31];
        for (int x : nums) {
            for (int i = 0; i < 31; ++i) {
                if ((x >> i & 1) == 1) {
                    ++cnt[i];
                }
            }
        }
        long ans = 0;
        while (k-- > 0) {
            int x = 0;
            for (int i = 0; i < 31; ++i) {
                if (cnt[i] > 0) {
                    x |= 1 << i;
                    --cnt[i];
                }
            }
            ans = (ans + 1L * x * x) % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
func maxSum(nums []int, k int) (ans int) {
	cnt := [31]int{}
	for _, x := range nums {
		for i := 0; i < 31; i++ {
			if x>>i&1 == 1 {
				cnt[i]++
			}
		}
	}
	const mod int = 1e9 + 7
	for ; k > 0; k-- {
		x := 0
		for i := 0; i < 31; i++ {
			if cnt[i] > 0 {
				x |= 1 << i
				cnt[i]--
			}
		}
		ans = (ans + x*x) % mod
	}
	return
}

# Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
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

// Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
// class Solution {
//     public int maxSum(List<Integer> nums, int k) {
//         final int mod = (int) 1e9 + 7;
//         int[] cnt = new int[31];
//         for (int x : nums) {
//             for (int i = 0; i < 31; ++i) {
//                 if ((x >> i & 1) == 1) {
//                     ++cnt[i];
//                 }
//             }
//         }
//         long ans = 0;
//         while (k-- > 0) {
//             int x = 0;
//             for (int i = 0; i < 31; ++i) {
//                 if (cnt[i] > 0) {
//                     x |= 1 << i;
//                     --cnt[i];
//                 }
//             }
//             ans = (ans + 1L * x * x) % mod;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2897: Apply Operations on Array to Maximize Sum of Squares
function maxSum(nums: number[], k: number): number {
    const cnt: number[] = Array(31).fill(0);
    for (const x of nums) {
        for (let i = 0; i < 31; ++i) {
            if ((x >> i) & 1) {
                ++cnt[i];
            }
        }
    }
    let ans = 0n;
    const mod = 1e9 + 7;
    while (k-- > 0) {
        let x = 0;
        for (let i = 0; i < 31; ++i) {
            if (cnt[i] > 0) {
                x |= 1 << i;
                --cnt[i];
            }
        }
        ans = (ans + BigInt(x) * BigInt(x)) % BigInt(mod);
    }
    return Number(ans);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(log M)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.