LeetCode #1144 — MEDIUM

Decrease Elements To Make Array Zigzag

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.

An array A is a zigzag array if either:

Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ...
OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ...

Return the minimum number of moves to transform the given array nums into a zigzag array.

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: We can decrease 2 to 0 or 3 to 1.

Example 2:

Input: nums = [9,6,1,6,2]
Output: 4

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: Given an array nums of integers, a move consists of choosing any element and decreasing it by 1. An array A is a zigzag array if either: Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ... OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ... Return the minimum number of moves to transform the given array nums into a zigzag array.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,2,3]

Example 2

[9,6,1,6,2]

Step 02

Core Insight

What unlocks the optimal approach

Do each case (even indexed is greater, odd indexed is greater) separately. In say the even case, you should decrease each even-indexed element until it is lower than its immediate neighbors.

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

Upper-end input sizes

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 #1144: Decrease Elements To Make Array Zigzag
class Solution {
    public int movesToMakeZigzag(int[] nums) {
        int[] ans = new int[2];
        int n = nums.length;
        for (int i = 0; i < 2; ++i) {
            for (int j = i; j < n; j += 2) {
                int d = 0;
                if (j > 0) {
                    d = Math.max(d, nums[j] - nums[j - 1] + 1);
                }
                if (j < n - 1) {
                    d = Math.max(d, nums[j] - nums[j + 1] + 1);
                }
                ans[i] += d;
            }
        }
        return Math.min(ans[0], ans[1]);
    }
}

// Accepted solution for LeetCode #1144: Decrease Elements To Make Array Zigzag
func movesToMakeZigzag(nums []int) int {
	ans := [2]int{}
	n := len(nums)
	for i := 0; i < 2; i++ {
		for j := i; j < n; j += 2 {
			d := 0
			if j > 0 {
				d = max(d, nums[j]-nums[j-1]+1)
			}
			if j < n-1 {
				d = max(d, nums[j]-nums[j+1]+1)
			}
			ans[i] += d
		}
	}
	return min(ans[0], ans[1])
}

# Accepted solution for LeetCode #1144: Decrease Elements To Make Array Zigzag
class Solution:
    def movesToMakeZigzag(self, nums: List[int]) -> int:
        ans = [0, 0]
        n = len(nums)
        for i in range(2):
            for j in range(i, n, 2):
                d = 0
                if j:
                    d = max(d, nums[j] - nums[j - 1] + 1)
                if j < n - 1:
                    d = max(d, nums[j] - nums[j + 1] + 1)
                ans[i] += d
        return min(ans)

// Accepted solution for LeetCode #1144: Decrease Elements To Make Array Zigzag
struct Solution;

impl Solution {
    fn moves_to_make_zigzag(nums: Vec<i32>) -> i32 {
        let mut sums = vec![0, 0];
        let n = nums.len();
        for i in 0..n {
            let mut adj = vec![];
            if i > 0 {
                adj.push(nums[i - 1]);
            }
            if i + 1 < n {
                adj.push(nums[i + 1]);
            }
            let min = adj.into_iter().min().unwrap();
            if nums[i] >= min {
                sums[i % 2] += (nums[i] - min) + 1;
            }
        }
        sums[0].min(sums[1])
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 3];
    let res = 2;
    assert_eq!(Solution::moves_to_make_zigzag(nums), res);
    let nums = vec![9, 6, 1, 6, 2];
    let res = 4;
    assert_eq!(Solution::moves_to_make_zigzag(nums), res);
    let nums = vec![10, 4, 4, 10, 10, 6, 2, 3];
    let res = 13;
    assert_eq!(Solution::moves_to_make_zigzag(nums), res);
}

// Accepted solution for LeetCode #1144: Decrease Elements To Make Array Zigzag
function movesToMakeZigzag(nums: number[]): number {
    const ans: number[] = Array(2).fill(0);
    const n = nums.length;
    for (let i = 0; i < 2; ++i) {
        for (let j = i; j < n; j += 2) {
            let d = 0;
            if (j > 0) {
                d = Math.max(d, nums[j] - nums[j - 1] + 1);
            }
            if (j < n - 1) {
                d = Math.max(d, nums[j] - nums[j + 1] + 1);
            }
            ans[i] += d;
        }
    }
    return Math.min(...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 n)

Space

O(1)

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.

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.