LeetCode #1716 — EASY

Calculate Money in Leetcode Bank

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

Hercy wants to save money for his first car. He puts money in the Leetcode bank every day.

He starts by putting in $1 on Monday, the first day. Every day from Tuesday to Sunday, he will put in $1 more than the day before. On every subsequent Monday, he will put in $1 more than the previous Monday.

Given n, return the total amount of money he will have in the Leetcode bank at the end of the n^th day.

Example 1:

Input: n = 4
Output: 10
Explanation: After the 4^th day, the total is 1 + 2 + 3 + 4 = 10.

Example 2:

Input: n = 10
Output: 37
Explanation: After the 10^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4) = 37. Notice that on the 2^nd Monday, Hercy only puts in $2.

Example 3:

Input: n = 20
Output: 96
Explanation: After the 20^th day, the total is (1 + 2 + 3 + 4 + 5 + 6 + 7) + (2 + 3 + 4 + 5 + 6 + 7 + 8) + (3 + 4 + 5 + 6 + 7 + 8) = 96.

Constraints:

1 <= n <= 1000

Step 01

Brute Force Baseline

Problem summary: Hercy wants to save money for his first car. He puts money in the Leetcode bank every day. He starts by putting in $1 on Monday, the first day. Every day from Tuesday to Sunday, he will put in $1 more than the day before. On every subsequent Monday, he will put in $1 more than the previous Monday. Given n, return the total amount of money he will have in the Leetcode bank at the end of the nth day.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

Simulate the process by keeping track of how much money Hercy is putting in and which day of the week it is, and use this information to deduce how much money Hercy will put in the next day.

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 #1716: Calculate Money in Leetcode Bank
class Solution {
    public int totalMoney(int n) {
        int k = n / 7, b = n % 7;
        int s1 = (28 + 28 + 7 * (k - 1)) * k / 2;
        int s2 = (k + 1 + k + 1 + b - 1) * b / 2;
        return s1 + s2;
    }
}

// Accepted solution for LeetCode #1716: Calculate Money in Leetcode Bank
func totalMoney(n int) int {
	k, b := n/7, n%7
	s1 := (28 + 28 + 7*(k-1)) * k / 2
	s2 := (k + 1 + k + 1 + b - 1) * b / 2
	return s1 + s2
}

# Accepted solution for LeetCode #1716: Calculate Money in Leetcode Bank
class Solution:
    def totalMoney(self, n: int) -> int:
        k, b = divmod(n, 7)
        s1 = (28 + 28 + 7 * (k - 1)) * k // 2
        s2 = (k + 1 + k + 1 + b - 1) * b // 2
        return s1 + s2

// Accepted solution for LeetCode #1716: Calculate Money in Leetcode Bank
struct Solution;

impl Solution {
    fn total_money(n: i32) -> i32 {
        let mut res = 0;
        for i in 0..n {
            res += (i % 7) + (i / 7) + 1;
        }
        res
    }
}

#[test]
fn test() {
    let n = 4;
    let res = 10;
    assert_eq!(Solution::total_money(n), res);
    let n = 10;
    let res = 37;
    assert_eq!(Solution::total_money(n), res);
    let n = 20;
    let res = 96;
    assert_eq!(Solution::total_money(n), res);
}

// Accepted solution for LeetCode #1716: Calculate Money in Leetcode Bank
function totalMoney(n: number): number {
    const k = (n / 7) | 0;
    const b = n % 7;
    const s1 = (((28 + 28 + 7 * (k - 1)) * k) / 2) | 0;
    const s2 = (((k + 1 + k + 1 + b - 1) * b) / 2) | 0;
    return s1 + s2;
}

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(1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.