LeetCode #1169 — MEDIUM

Invalid Transactions

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A transaction is possibly invalid if:

the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.

You are given an array of strings transaction where transactions[i] consists of comma-separated values representing the name, time (in minutes), amount, and city of the transaction.

Return a list of transactions that are possibly invalid. You may return the answer in any order.

Example 1:

Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.

Example 2:

Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]

Example 3:

Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]

Constraints:

transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000.

Step 01

Brute Force Baseline

Problem summary: A transaction is possibly invalid if: the amount exceeds $1000, or; if it occurs within (and including) 60 minutes of another transaction with the same name in a different city. You are given an array of strings transaction where transactions[i] consists of comma-separated values representing the name, time (in minutes), amount, and city of the transaction. Return a list of transactions that are possibly invalid. You may return the answer in any order.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["alice,20,800,mtv","alice,50,100,beijing"]

Example 2

["alice,20,800,mtv","alice,50,1200,mtv"]

Example 3

["alice,20,800,mtv","bob,50,1200,mtv"]

Step 02

Core Insight

What unlocks the optimal approach

Split each string into four arrays.
For each transaction check if it's invalid, you can do this with just a loop with help of the four arrays generated on step 1.
At the end you perform O(N ^ 2) operations.

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 #1169: Invalid Transactions
class Solution {
    public List<String> invalidTransactions(String[] transactions) {
        Map<String, List<Item>> d = new HashMap<>();
        Set<Integer> idx = new HashSet<>();
        for (int i = 0; i < transactions.length; ++i) {
            var e = transactions[i].split(",");
            String name = e[0];
            int time = Integer.parseInt(e[1]);
            int amount = Integer.parseInt(e[2]);
            String city = e[3];
            d.computeIfAbsent(name, k -> new ArrayList<>()).add(new Item(time, city, i));
            if (amount > 1000) {
                idx.add(i);
            }
            for (Item item : d.get(name)) {
                if (!city.equals(item.city) && Math.abs(time - item.t) <= 60) {
                    idx.add(i);
                    idx.add(item.i);
                }
            }
        }
        List<String> ans = new ArrayList<>();
        for (int i : idx) {
            ans.add(transactions[i]);
        }
        return ans;
    }
}

class Item {
    int t;
    String city;
    int i;

    Item(int t, String city, int i) {
        this.t = t;
        this.city = city;
        this.i = i;
    }
}

// Accepted solution for LeetCode #1169: Invalid Transactions
func invalidTransactions(transactions []string) (ans []string) {
	d := map[string][]tuple{}
	idx := map[int]bool{}
	for i, x := range transactions {
		e := strings.Split(x, ",")
		name := e[0]
		time, _ := strconv.Atoi(e[1])
		amount, _ := strconv.Atoi(e[2])
		city := e[3]
		d[name] = append(d[name], tuple{time, city, i})
		if amount > 1000 {
			idx[i] = true
		}
		for _, item := range d[name] {
			if city != item.city && abs(time-item.t) <= 60 {
				idx[i], idx[item.i] = true, true
			}
		}
	}
	for i := range idx {
		ans = append(ans, transactions[i])
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

type tuple struct {
	t    int
	city string
	i    int
}

# Accepted solution for LeetCode #1169: Invalid Transactions
class Solution:
    def invalidTransactions(self, transactions: List[str]) -> List[str]:
        d = defaultdict(list)
        idx = set()
        for i, x in enumerate(transactions):
            name, time, amount, city = x.split(",")
            time, amount = int(time), int(amount)
            d[name].append((time, city, i))
            if amount > 1000:
                idx.add(i)
            for t, c, j in d[name]:
                if c != city and abs(time - t) <= 60:
                    idx.add(i)
                    idx.add(j)
        return [transactions[i] for i in idx]

// Accepted solution for LeetCode #1169: Invalid Transactions
struct Solution;
use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn invalid_transactions(transactions: Vec<String>) -> Vec<String> {
        let n = transactions.len();
        let mut hm: HashMap<String, Vec<(i32, String, String)>> = HashMap::new();
        let mut hs: HashSet<String> = HashSet::new();
        for i in 0..n {
            let transaction = &transactions[i];
            let mut v: Vec<String> = transaction
                .split_terminator(',')
                .map(|s| s.to_string())
                .collect();
            let city = v.pop().unwrap();
            let amount = v.pop().unwrap().parse::<i32>().unwrap();
            let time = v.pop().unwrap().parse::<i32>().unwrap();
            let name = v.pop().unwrap();
            hm.entry(name)
                .or_default()
                .push((time, city, transaction.clone()));
            if amount > 1000 {
                hs.insert(transactions[i].clone());
            }
        }
        for transactions in hm.values() {
            let n = transactions.len();
            for i in 0..n {
                for j in i + 1..n {
                    let ti = &transactions[i];
                    let tj = &transactions[j];
                    if (ti.0 - tj.0).abs() <= 60 && ti.1 != tj.1 {
                        hs.insert(ti.2.clone());
                        hs.insert(tj.2.clone());
                    }
                }
            }
        }
        hs.drain().collect()
    }
}

#[test]
fn test() {
    let transactions = vec_string!["alice,20,800,mtv", "alice,50,100,beijing"];
    let mut res = vec_string!["alice,20,800,mtv", "alice,50,100,beijing"];
    let mut ans = Solution::invalid_transactions(transactions);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let transactions = vec_string!["alice,20,800,mtv", "alice,50,1200,mtv"];
    let mut res = vec_string!["alice,50,1200,mtv"];
    let mut ans = Solution::invalid_transactions(transactions);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
    let transactions = vec_string!["alice,20,800,mtv", "bob,50,1200,mtv"];
    let mut res = vec_string!["bob,50,1200,mtv"];
    let mut ans = Solution::invalid_transactions(transactions);
    res.sort();
    ans.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #1169: Invalid Transactions
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1169: Invalid Transactions
// class Solution {
//     public List<String> invalidTransactions(String[] transactions) {
//         Map<String, List<Item>> d = new HashMap<>();
//         Set<Integer> idx = new HashSet<>();
//         for (int i = 0; i < transactions.length; ++i) {
//             var e = transactions[i].split(",");
//             String name = e[0];
//             int time = Integer.parseInt(e[1]);
//             int amount = Integer.parseInt(e[2]);
//             String city = e[3];
//             d.computeIfAbsent(name, k -> new ArrayList<>()).add(new Item(time, city, i));
//             if (amount > 1000) {
//                 idx.add(i);
//             }
//             for (Item item : d.get(name)) {
//                 if (!city.equals(item.city) && Math.abs(time - item.t) <= 60) {
//                     idx.add(i);
//                     idx.add(item.i);
//                 }
//             }
//         }
//         List<String> ans = new ArrayList<>();
//         for (int i : idx) {
//             ans.add(transactions[i]);
//         }
//         return ans;
//     }
// }
// 
// class Item {
//     int t;
//     String city;
//     int i;
// 
//     Item(int t, String city, int i) {
//         this.t = t;
//         this.city = city;
//         this.i = i;
//     }
// }

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(n^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.