LeetCode 3674 - Minimum Operations to Equalize Array

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leetcodeeasyarraybit-manipulationbrainteaser

LeetCode Problem 3674

Difficulty: 🟢 Easy
Topics: Array, Bit Manipulation, Brainteaser

Solution

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Problem Understanding

The problem asks us to make all elements of an integer array nums equal using the minimum number of operations. Each operation consists of selecting a contiguous subarray and replacing every element in that subarray with the bitwise AND of all elements in it.

The input is an array of integers nums of length n where 1 <= n <= 100 and 1 <= nums[i] <= 10^5. The output is a single integer representing the minimum number of operations needed to make all elements identical.

The constraints indicate that the array is small (at most 100 elements), so even O(n^2) algorithms may be acceptable, but we should strive for something linear or linearithmic if possible. The problem guarantees all numbers are positive integers, so the AND operation will not introduce negative values.

Important edge cases include arrays where all elements are already equal (requiring zero operations), arrays of length one, and arrays where the bitwise AND of all elements produces zero.

Approaches

Brute Force

A naive approach is to try every possible subarray, perform the bitwise AND, and recursively continue until all elements are equal. This approach guarantees correctness because it explores all sequences of operations, but it is extremely inefficient. For an array of length n, the number of possible subarrays is O(n^2), and the number of recursive sequences grows exponentially. Even with memoization, this is overkill for the given constraints.

Optimal Approach

The key insight is that the bitwise AND operation is monotonic and idempotent. If we compute the AND of the entire array, this is the only value that can ever be achieved across all elements. Once we know this target value, any segment of the array that already equals this value does not need modification.

The problem then reduces to counting the number of contiguous segments in nums that do not equal the target AND. Each such segment requires exactly one operation because we can select the whole segment at once. This insight transforms a complex recursive problem into a simple linear scan.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n^2) Explore all subarrays recursively
Optimal O(n) O(1) Scan array once to count segments not equal to AND of all elements

Algorithm Walkthrough

  1. Compute the bitwise AND of all elements in nums. This value is the target value to which all elements must be converted.
  2. Initialize a counter for operations (ops) and a boolean flag (in_segment) to track whether we are inside a contiguous segment that does not match the target.
  3. Iterate over each element of the array.
  4. If the current element equals the target, mark in_segment as false since we are outside a segment needing modification.
  5. If the current element does not equal the target and in_segment is false, this marks the start of a new segment. Increment ops by 1 and set in_segment to true.
  6. Continue until the end of the array. The counter ops will now contain the minimum number of operations required.

Why it works: The bitwise AND operation is idempotent and cannot create a value larger than the operands. The global AND of all elements is the smallest value that all elements can converge to. Any contiguous elements not already equal to this target can be made equal in a single operation, so counting segments yields the minimal operations.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        # Step 1: compute the AND of the entire array
        target = nums[0]
        for num in nums[1:]:
            target &= num
        
        # Step 2: count segments not equal to target
        ops = 0
        in_segment = False
        for num in nums:
            if num == target:
                in_segment = False
            else:
                if not in_segment:
                    ops += 1
                    in_segment = True
        return ops

This implementation first determines the smallest achievable value (target) by AND-ing all elements. It then counts the number of contiguous segments that differ from this target, ensuring minimal operations. The use of a flag avoids double-counting overlapping segments.

Go Solution

func minOperations(nums []int) int {
    n := len(nums)
    if n == 0 {
        return 0
    }

    // Step 1: compute the AND of the entire array
    target := nums[0]
    for i := 1; i < n; i++ {
        target &= nums[i]
    }

    // Step 2: count segments not equal to target
    ops := 0
    inSegment := false
    for i := 0; i < n; i++ {
        if nums[i] == target {
            inSegment = false
        } else {
            if !inSegment {
                ops++
                inSegment = true
            }
        }
    }
    return ops
}

The Go implementation mirrors the Python approach, using slices and a boolean flag. Handling an empty array is trivial, although the constraints guarantee n >= 1. Slice iteration replaces Python's for-each loop.

Worked Examples

Example 1: nums = [1,2]

Compute target AND: 1 & 2 = 0.

Iterate:

Index Value Target? In Segment Ops
0 1 No False -> True 1
1 2 No True 1

Result: ops = 1.

Example 2: nums = [5,5,5]

Compute target AND: 5 & 5 & 5 = 5.

Iterate:

Index Value Target? In Segment Ops
0 5 Yes False 0
1 5 Yes False 0
2 5 Yes False 0

Result: ops = 0.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass to compute AND and another pass to count segments
Space O(1) Constant extra space for counter and flag

This linear complexity is sufficient given n <= 100.

Test Cases

# Provided examples
assert Solution().minOperations([1,2]) == 1  # requires one operation
assert Solution().minOperations([5,5,5]) == 0  # all elements already equal

# Edge cases
assert Solution().minOperations([1]) == 0  # single element
assert Solution().minOperations([7,7,3,7]) == 1  # one segment needing AND
assert Solution().minOperations([1,3,7,1,3,7]) == 3  # multiple segments
assert Solution().minOperations([8,8,8,0,8,8]) == 1  # zero creates a single segment
Test Why
[1,2] Minimal example requiring a single operation
[5,5,5] Already uniform array
[1] Single-element array
[7,7,3,7] One segment differing from target
[1,3,7,1,3,7] Multiple segments differing
[8,8,8,0,8,8] Zero introduces a segment

Edge Cases

One important edge case is a single-element array. Since the array is already uniform, the algorithm correctly returns zero operations.

Another edge case is when all elements are the same except for a single element in the middle. The algorithm counts this as a single contiguous segment, correctly identifying one operation.

A third edge case involves multiple alternating segments, where several contiguous groups differ from the target. The algorithm correctly counts each separate segment, ensuring no unnecessary operations are added.

The method also gracefully handles zeros or large numbers due to the properties of the bitwise AND operation, which never produces values larger than its operands.

This completes a comprehensive, reference-style solution for LeetCode 3674, including Python and Go implementations, worked examples, complexity analysis, and edge case discussion.