Find Triangular Sum of an Array

Find Triangular Sum of an Array - Problem

You are given a 0-indexed integer array nums, where nums[i] is a digit between 0 and 9 (inclusive).

The triangular sum of nums is the value of the only element present in nums after the following process terminates:

Let nums comprise of n elements. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n - 1.
For each index i, where 0 <= i < n - 1, assign the value of newNums[i] as (nums[i] + nums[i+1]) % 10, where % denotes modulo operator.
Replace the array nums with newNums.
Repeat the entire process starting from step 1.

Return the triangular sum of nums.

Input & Output

Example 1 — Basic Case

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

› Output: 8

💡 Note: Layer by layer: [1,2,3,4,5] → [3,5,7,9] → [8,2,6] → [0,8] → [8]. The triangular sum is 8.

Example 2 — Small Array

$ Input: nums = [5]

› Output: 5

💡 Note: Single element array: the triangular sum is the element itself, which is 5.

Example 3 — Two Elements

$ Input: nums = [7,9]

› Output: 6

💡 Note: Two elements: (7 + 9) % 10 = 16 % 10 = 6. The triangular sum is 6.

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 9

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15

The key insight is recognizing the triangular sum follows Pascal's triangle pattern. The brute force approach simulates layer by layer in O(n²) time. The optimal mathematical approach uses binomial coefficients to calculate directly in O(n) time. Best approach: Mathematical formula with Time: O(n), Space: O(1).

Common Approaches

✓ Greedy

⏱️ Time: N/A Space: N/A

Brute Force Simulation

⏱️ Time: O(n²) Space: O(n²)

Directly follow the problem description: repeatedly create new arrays where each element is the sum of adjacent elements modulo 10, until only one element remains.

In-Place Simulation

⏱️ Time: O(n²) Space: O(1)

Instead of creating new arrays at each step, we can reuse the same array and just track the current length. This reduces space complexity significantly while maintaining the same logic.

Mathematical Formula (Pascal's Triangle)

⏱️ Time: O(n) Space: O(1)

The triangular sum follows Pascal's triangle pattern where each position contributes according to binomial coefficient C(n-1, i). We can calculate this directly using mathematical properties.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int triangularSum(int* nums, int numsSize) {
    if (numsSize == 1) {
        return nums[0];
    }
    
    int* current = malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        current[i] = nums[i];
    }
    
    int currentSize = numsSize;
    
    while (currentSize > 1) {
        int* next = malloc((currentSize - 1) * sizeof(int));
        for (int i = 0; i < currentSize - 1; i++) {
            next[i] = (current[i] + current[i + 1]) % 10;
        }
        free(current);
        current = next;
        currentSize--;
    }
    
    int result = current[0];
    free(current);
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int* nums = malloc(1000 * sizeof(int));
    int numsSize = 0;
    
    // Parse array [1,2,3,4,5]
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++; // Find opening bracket
    if (*ptr == '[') ptr++; // Skip opening bracket
    
    while (*ptr && *ptr != ']') {
        if (*ptr >= '0' && *ptr <= '9') {
            int num = 0;
            while (*ptr >= '0' && *ptr <= '9') {
                num = num * 10 + (*ptr - '0');
                ptr++;
            }
            nums[numsSize++] = num;
        } else {
            ptr++;
        }
    }
    
    int result = triangularSum(nums, numsSize);
    printf("%d\n", result);
    
    free(nums);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚠ Quadratic Space

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