Non Overlapping Intervals Minimum Removal in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimum removals for overlapping intervals

What is the output of the following code that calculates the minimum number of intervals to remove to avoid overlaps?

DSA C

int intervals[][2] = {{1,3},{2,4},{3,5},{6,7}};
int n = 4;

// Sort intervals by end time
for(int i=0; i<n-1; i++) {
    for(int j=0; j<n-i-1; j++) {
        if(intervals[j][1] > intervals[j+1][1]) {
            int temp0 = intervals[j][0];
            int temp1 = intervals[j][1];
            intervals[j][0] = intervals[j+1][0];
            intervals[j][1] = intervals[j+1][1];
            intervals[j+1][0] = temp0;
            intervals[j+1][1] = temp1;
        }
    }
}

int count = 1;
int end = intervals[0][1];
for(int i=1; i<n; i++) {
    if(intervals[i][0] >= end) {
        count++;
        end = intervals[i][1];
    }
}

int result = n - count;
printf("%d\n", result);

A 1

B 2

C 3

D 0

💡 Hint

Think about how many intervals can fit without overlapping after sorting by end time.

The code sorts intervals by their end time and greedily counts the maximum number of non-overlapping intervals. The minimum removals is total intervals minus this count. Here, 3 intervals can be non-overlapping, so 1 must be removed.

🧠 Conceptual

intermediate

Understanding the greedy choice in interval removal

Why does sorting intervals by their end time help in minimizing the number of intervals to remove to avoid overlaps?

A Because intervals with earliest start time always overlap less.

B Because sorting by length of intervals guarantees minimal removals.

C Because sorting by start time ensures maximum intervals are removed.

D Because choosing intervals with earliest end time leaves more room for following intervals.

💡 Hint

Think about how the end time affects the possibility of fitting more intervals after it.

Choosing intervals that end earliest allows more intervals to fit after them without overlapping, minimizing removals.

🔧 Debug

advanced

Identify the bug in interval sorting code

What is the bug in the following code snippet that sorts intervals by their end time? for(int i=0; i intervals[j+1][1]) { int temp0 = intervals[j][0]; int temp1 = intervals[j][1]; intervals[j][0] = intervals[j+1][0]; intervals[j][1] = intervals[j+1][1]; intervals[j+1][0] = temp0; intervals[j+1][1] = temp1; } } }

A Swapping intervals incorrectly swaps only end times, not start times.

B The inner loop should run until j < n-i, not n-i-1.

C No bug, the code correctly sorts intervals by end time.

D The swap logic is correct but inefficient; no bug affects correctness.

💡 Hint

Check carefully how the swap is done for both start and end times.

The code correctly swaps both start and end times of intervals to sort by end time using bubble sort.

❓ Predict Output

advanced

Output of minimum removals with overlapping intervals

What is the output of the following code snippet?

DSA C

int intervals[][2] = {{1,2},{2,3},{3,4},{1,3}};
int n = 4;

// Sort intervals by end time
for(int i=0; i<n-1; i++) {
    for(int j=0; j<n-i-1; j++) {
        if(intervals[j][1] > intervals[j+1][1]) {
            int temp0 = intervals[j][0];
            int temp1 = intervals[j][1];
            intervals[j][0] = intervals[j+1][0];
            intervals[j][1] = intervals[j+1][1];
            intervals[j+1][0] = temp0;
            intervals[j+1][1] = temp1;
        }
    }
}

int count = 1;
int end = intervals[0][1];
for(int i=1; i<n; i++) {
    if(intervals[i][0] >= end) {
        count++;
        end = intervals[i][1];
    }
}

int result = n - count;
printf("%d\n", result);

A 1

B 2

C 3

D 0

💡 Hint

Count how many intervals can be chosen without overlap after sorting by end time.

The code sorts intervals by their end time and greedily counts the maximum number of non-overlapping intervals. The minimum removals is total intervals minus this count. Here, 3 intervals can be non-overlapping, so 1 must be removed.

🧠 Conceptual

expert

Why greedy approach works for minimum interval removals

Which statement best explains why the greedy approach of selecting intervals by earliest end time guarantees the minimum number of removals to avoid overlaps?

A Because it tries all possible combinations of intervals to find the minimum removals.

B Because selecting intervals with earliest end time always leads to the globally optimal solution by maximizing remaining space.

C Because it removes the longest intervals first to minimize overlap.

D Because it sorts intervals by start time and removes those that start too early.

💡 Hint

Think about how the greedy choice leaves room for future intervals.

Choosing intervals with earliest end time leaves the maximum room for subsequent intervals, ensuring the maximum number of intervals remain without overlap, thus minimizing removals.