Array Prototype Last

Array Prototype Last - Problem

JavaScript Easy

Write code that enhances all arrays such that you can call the array.last() method on any array and it will return the last element. If there are no elements in the array, it should return -1.

You may assume the array is the output of JSON.parse.

Note: This problem requires extending JavaScript's Array prototype with a custom method.

Input & Output

Example 1 — Basic Array

$ Input: arr = [1, 2, 3, 4, 5]

› Output: 5

💡 Note: The array has 5 elements, so the last element at index 4 is 5

Example 2 — Single Element

$ Input: arr = [10]

› Output: 10

💡 Note: Array with one element - the first and last element are the same

Example 3 — Empty Array

$ Input: arr = []

› Output: -1

💡 Note: Empty array has no elements, so return -1 as specified

Constraints

0 ≤ arr.length ≤ 1000
-1000 ≤ arr[i] ≤ 1000

Visualization

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Asked in

G Google 25 a Amazon 20 f Facebook 15

The key insight is that we can use the array's length property to directly access the last element at index length - 1. The optimal approach uses direct indexing with O(1) time and O(1) space complexity.

Common Approaches

✓ Binary Search

⏱️ Time: N/A Space: N/A

Direct Index Access

⏱️ Time: O(1) Space: O(1)

Instead of iterating, use the array's length property to calculate the index of the last element and access it directly. This is the most efficient approach.

Linear Scan

⏱️ Time: O(n) Space: O(1)

Loop through all elements from start to finish, keeping track of each element until we reach the end. This simulates not knowing the array length.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static int nums[1000];

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int numsSize = 0;
    
    // Parse the array from input like [1,2,3,4,5]
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++; // Find opening bracket
    if (*ptr == '[') ptr++; // Skip opening bracket
    
    // Parse numbers until we hit ']'
    while (*ptr && *ptr != ']') {
        // Skip whitespace and commas
        while (*ptr == ' ' || *ptr == ',') ptr++;
        
        // If we hit the closing bracket or end, break
        if (*ptr == ']' || *ptr == '\0') break;
        
        // Parse the number
        char* endptr;
        int num = (int)strtol(ptr, &endptr, 10);
        if (endptr != ptr) { // Successfully parsed a number
            nums[numsSize++] = num;
            ptr = endptr;
        } else {
            ptr++; // Skip invalid character
        }
    }
    
    // Return last element or -1 if empty
    if (numsSize == 0) {
        printf("-1\n");
    } else {
        printf("%d\n", nums[numsSize - 1]);
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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