Count Iterations of Consecutive Loops

Easy5 min

Topics

BasicsComplexity

Companies

TCSInfosysWipro

A function runs three loops one after another: one over n, one over m, and another over n. Given n and m, return the total number of loop passes. When loops are sequential, their work adds up: n + m + n = 2n + m, which is O(n + m). Contrast this with nested loops, where counts multiply. Telling "one after another" from "one inside another" is most of the skill in reading complexity from code.

Test Case 1:

Input:n = 3, m = 5

Output:11

Explanation:The first loop runs 3 times, the second 5, the third 3 again: 3 + 5 + 3.

Test Case 2:

Input:n = 0, m = 4

Output:4

Explanation:The loops over n never run; only the m loop does.

Constraints

◆0 ≤ n, m ≤ 1000000

Try the Dry Run

Mode

Approach

Approach & Solutions

Run All Three LoopsGood

Run the three loops one after another — n passes, then m passes, then n passes again — and count every pass. Because each loop finishes before the next starts, their counts add up: 2n + m.

TimeO(n + m)

SpaceO(1)

Java

Try it out

class Solution {
    public int consecutiveLoopsCount(int n, int m) {
        int count = 0;
        for (int i = 0; i < n; i++) {
            count++;
        }
        for (int j = 0; j < m; j++) {
            count++;
        }
        for (int i = 0; i < n; i++) {
            count++;
        }
        return count;
    }
}

Add the CountsOptimal

Each loop's count is known without running it: n, m and n. Add them: 2n + m. In Big-O terms the constant 2 disappears and the code is O(n + m) — both inputs matter, because either one could be the big one.

TimeO(1)