Nested loops in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested loops with break

What is the output of this C code snippet?

C

#include <stdio.h>

int main() {
    for (int i = 1; i <= 3; i++) {
        for (int j = 1; j <= 3; j++) {
            if (j == 2) break;
            printf("%d%d ", i, j);
        }
    }
    return 0;
}

A 11 12 13 21 22 23 31 32 33

B 11 21 31

C 11 12 21 22 31 32

D 11 21 22 31 32

💡 Hint

Remember that break exits the inner loop immediately when j == 2.

The inner loop prints i and j until j == 2, then breaks. So only when j == 1 the print happens for each i.

❓ Predict Output

intermediate

Counting iterations in nested loops

What is the value of count after running this code?

C

#include <stdio.h>

int main() {
    int count = 0;
    for (int i = 0; i < 4; i++) {
        for (int j = i; j < 4; j++) {
            count++;
        }
    }
    printf("%d", count);
    return 0;
}

A 12

B 16

C 8

D 10

💡 Hint

Count how many times the inner loop runs for each i.

When i=0, inner loop runs 4 times; i=1 runs 3 times; i=2 runs 2 times; i=3 runs 1 time. Total = 4+3+2+1 = 10.

🔧 Debug

advanced

Identify the error in nested loops

What error does this code produce when compiled?

C

#include <stdio.h>

int main() {
    for (int i = 0; i < 3; i++)
        for (int j = 0; j < 3; j++)
            printf("%d %d\n", i, j);
    printf("Done\n");
    return 0;
}

A Compilation error: expected ';' before 'printf'

B No error, prints pairs and 'Done' three times

C No error, prints pairs and 'Done' once

D Runtime error: segmentation fault

💡 Hint

Check how the indentation affects which statements belong to the loops.

Only the inner loop is controlled by the second for. The printf("Done\n") is outside both loops and runs once after all loops finish.

🧠 Conceptual

advanced

Number of iterations in nested loops with conditions

What value does this program print?

C

#include <stdio.h>

int main() {
    int count = 0;
    for (int i = 1; i <= 5; i++) {
        for (int j = 1; j <= 5; j++) {
            if (i * j % 2 == 0) {
                count++;
            }
        }
    }
    printf("%d", count);
    return 0;
}

A 16

B 10

C 25

D 15

💡 Hint

Count pairs (i, j) where product is even.

5x5=25 total pairs. i * j % 2 == 0 when at least one is even. Odds: 1,3,5 (3); evens: 2,4 (2). Odd-odd pairs (odd product): 3x3=9. Even products: 25-9=16.

❓ Predict Output

expert

Output of nested loops with continue and multiple conditions

What is the output of this code?

C

#include <stdio.h>

int main() {
    for (int i = 1; i <= 3; i++) {
        for (int j = 1; j <= 3; j++) {
            if (i == j) continue;
            if (i + j == 4) continue;
            printf("%d%d ", i, j);
        }
    }
    return 0;
}

A 12 21 23 32

B 12 13 21 31 32

C 13 21 31 32

D 12 13 21 23 31 32

💡 Hint

Skip pairs where i == j or i + j == 4.

Skips when i == j (11,22,33) or i + j == 4 (13,22,31). Printed: 12, 21, 23, 32.