Decrease Elements To Make Array Zigzag

Decrease Elements To Make Array Zigzag - Problem

Array Greedy Medium

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

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,17,5]

› Output: 0

💡 Note: Array [1,17,5] already follows zigzag pattern with odd indices as peaks: 1 < 17 > 5. No moves needed.

Example 2 — All Equal

$ Input: nums = [2,2,2]

› Output: 1

💡 Note: Make middle element peak: keep [2,2,2] but need one valley. Best is [1,2,1] requiring 2 moves, or [2,2,1] requiring 1 move for pattern with even peaks.

Example 3 — Already Zigzag

$ Input: nums = [1,3,2,4,1]

› Output: 0

💡 Note: Array already follows zigzag pattern with odd indices as peaks: 1 < 3 > 2 < 4 > 1. No moves needed.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that there are only two possible zigzag patterns: even-indexed peaks or odd-indexed peaks. We calculate moves for both patterns and choose the minimum. Best approach is greedy two-pattern comparison. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Try All Combinations	O(2^n)	O(n)	Generate all possible zigzag arrays and find minimum moves
Greedy - Two Pattern Approach	O(n)	O(1)	Try both zigzag patterns separately and choose minimum moves

Brute Force - Try All Combinations — Algorithm Steps

Generate all possible zigzag arrays
Calculate moves for each configuration
Return minimum moves found

Visualization

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Step-by-Step Walkthrough

Generate

Create all possible zigzag arrays

Validate

Check if each array is valid zigzag

Optimize

Find minimum moves needed

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int minMoves = INT_MAX;

int isZigzag(int* arr, int size) {
    if (size <= 1) return 1;
    
    // Check pattern 1: even indices are peaks
    int pattern1 = 1;
    for (int i = 0; i < size; i++) {
        if (i % 2 == 0) {
            if ((i > 0 && arr[i] <= arr[i-1]) || (i < size-1 && arr[i] <= arr[i+1])) {
                pattern1 = 0;
                break;
            }
        } else {
            if ((i > 0 && arr[i] >= arr[i-1]) || (i < size-1 && arr[i] >= arr[i+1])) {
                pattern1 = 0;
                break;
            }
        }
    }
    
    if (pattern1) return 1;
    
    // Check pattern 2: odd indices are peaks
    int pattern2 = 1;
    for (int i = 0; i < size; i++) {
        if (i % 2 == 1) {
            if ((i > 0 && arr[i] <= arr[i-1]) || (i < size-1 && arr[i] <= arr[i+1])) {
                pattern2 = 0;
                break;
            }
        } else {
            if ((i > 0 && arr[i] >= arr[i-1]) || (i < size-1 && arr[i] >= arr[i+1])) {
                pattern2 = 0;
                break;
            }
        }
    }
    
    return pattern2;
}

void backtrack(int* nums, int numsSize, int index, int* current, int currentSize, int moves) {
    if (index == numsSize) {
        if (isZigzag(current, currentSize)) {
            if (moves < minMoves) {
                minMoves = moves;
            }
        }
        return;
    }
    
    int original = nums[index];
    for (int decrease = 0; decrease <= original; decrease++) {
        current[currentSize] = original - decrease;
        backtrack(nums, numsSize, index + 1, current, currentSize + 1, moves + decrease);
    }
}

int solution(int* nums, int numsSize) {
    if (numsSize <= 1) return 0;
    
    minMoves = INT_MAX;
    int* current = (int*)malloc(numsSize * sizeof(int));
    backtrack(nums, numsSize, 0, current, 0, 0);
    free(current);
    return minMoves == INT_MAX ? 0 : minMoves;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

For each element, we try multiple height values exponentially

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack and temporary arrays

⚡ Linearithmic Space

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Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

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Code -

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