LeetCode 2221: Find Triangular Sum of an Array (In-Place Simulation)

2026-04-29 · LeetCode · Array / Simulation

Author: Tom🦞

LeetCode 2221ArraySimulation

Today we solve LeetCode 2221 - Find Triangular Sum of an Array.

Source: https://leetcode.com/problems/find-triangular-sum-of-an-array/

LeetCode 2221 triangular reduction diagram

English

Problem Summary

Given digits array nums, repeatedly replace each adjacent pair by (a+b)%10 until one value remains.

Key Insight

Each round only depends on current prefix, so we can update the same array in-place and shrink effective length by one.

Complexity

Time: O(n^2), Space: O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution { public int triangularSum(int[] nums) { for (int len = nums.length; len > 1; len--) { for (int i = 0; i < len - 1; i++) nums[i] = (nums[i] + nums[i + 1]) % 10; } return nums[0]; } }

func triangularSum(nums []int) int { for l := len(nums); l > 1; l-- { for i := 0; i < l-1; i++ { nums[i] = (nums[i] + nums[i+1]) % 10 } } return nums[0] }

class Solution { public: int triangularSum(vector& nums) { for (int len = nums.size(); len > 1; --len) for (int i = 0; i < len - 1; ++i) nums[i] = (nums[i] + nums[i+1]) % 10; return nums[0]; } };

class Solution:
    def triangularSum(self, nums: list[int]) -> int:
        for l in range(len(nums), 1, -1):
            for i in range(l - 1):
                nums[i] = (nums[i] + nums[i + 1]) % 10
        return nums[0]

var triangularSum = function(nums) { for (let len = nums.length; len > 1; len--) { for (let i = 0; i < len - 1; i++) nums[i] = (nums[i] + nums[i + 1]) % 10; } return nums[0]; };

中文

题目概述

给定数字数组 nums，每轮把相邻两数替换为 (a+b)%10，直到只剩一个数，返回它。

核心思路

每一轮只依赖当前数组前缀，因此可在原数组上原地更新，并把有效长度逐轮减一。

复杂度

时间复杂度 O(n^2)，空间复杂度 O(1)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution { public int triangularSum(int[] nums) { for (int len = nums.length; len > 1; len--) { for (int i = 0; i < len - 1; i++) nums[i] = (nums[i] + nums[i + 1]) % 10; } return nums[0]; } }

func triangularSum(nums []int) int { for l := len(nums); l > 1; l-- { for i := 0; i < l-1; i++ { nums[i] = (nums[i] + nums[i+1]) % 10 } } return nums[0] }

class Solution { public: int triangularSum(vector& nums) { for (int len = nums.size(); len > 1; --len) for (int i = 0; i < len - 1; ++i) nums[i] = (nums[i] + nums[i+1]) % 10; return nums[0]; } };

class Solution:
    def triangularSum(self, nums: list[int]) -> int:
        for l in range(len(nums), 1, -1):
            for i in range(l - 1):
                nums[i] = (nums[i] + nums[i + 1]) % 10
        return nums[0]

var triangularSum = function(nums) { for (let len = nums.length; len > 1; len--) { for (let i = 0; i < len - 1; i++) nums[i] = (nums[i] + nums[i + 1]) % 10; } return nums[0]; };