Merge Overlapping Intervals in DSA C

Mental Model

We combine intervals that overlap into one bigger interval so no two intervals cover the same range twice.

Analogy: Imagine booking meeting rooms. If two meetings overlap in time, you merge them into one longer meeting to avoid confusion.

Intervals before merge:
[1,3] -> [2,6] -> [8,10] -> [15,18] -> null

Dry Run Walkthrough

Input: intervals: [1,3], [2,6], [8,10], [15,18]

Goal: Merge all overlapping intervals into combined intervals without overlaps

Step 1: Sort intervals by start time

[1,3] -> [2,6] -> [8,10] -> [15,18] -> null

Why: Sorting helps to easily find overlapping intervals by comparing neighbors

Step 2: Start with first interval [1,3] in merged list

Merged: [1,3]

Why: We need a base interval to compare others against

Step 3: Compare next interval [2,6] with last merged [1,3], they overlap

Merged: [1,6]

Why: Since 2 ≤ 3, intervals overlap, so merge by extending end to max(3,6)=6

Step 4: Compare next interval [8,10] with last merged [1,6], no overlap

Merged: [1,6] -> [8,10]

Why: 8 > 6 means no overlap, so add new interval

Step 5: Compare next interval [15,18] with last merged [8,10], no overlap

Merged: [1,6] -> [8,10] -> [15,18]

Why: 15 > 10 means no overlap, so add new interval

Result:

Merged intervals:
[1,6] -> [8,10] -> [15,18] -> null

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

// Define interval structure
typedef struct {
    int start;
    int end;
} Interval;

// Comparator for qsort to sort intervals by start time
int compare(const void *a, const void *b) {
    Interval *i1 = (Interval *)a;
    Interval *i2 = (Interval *)b;
    return i1->start - i2->start;
}

// Function to merge overlapping intervals
// intervals: input array, n: number of intervals
// merged: output array to store merged intervals
// returns number of merged intervals
int mergeIntervals(Interval intervals[], int n, Interval merged[]) {
    if (n == 0) return 0;

    // Sort intervals by start time
    qsort(intervals, n, sizeof(Interval), compare);

    int idx = 0; // index for merged array
    merged[0] = intervals[0]; // start with first interval

    for (int i = 1; i < n; i++) {
        // If current interval overlaps with last merged
        if (intervals[i].start <= merged[idx].end) {
            // Merge by extending the end if needed
            if (intervals[i].end > merged[idx].end) {
                merged[idx].end = intervals[i].end;
            }
        } else {
            // No overlap, add new interval
            idx++;
            merged[idx] = intervals[i];
        }
    }

    return idx + 1; // number of merged intervals
}

// Helper to print intervals
void printIntervals(Interval intervals[], int n) {
    for (int i = 0; i < n; i++) {
        printf("[%d,%d]", intervals[i].start, intervals[i].end);
        if (i != n - 1) printf(" -> ");
    }
    printf(" -> null\n");
}

int main() {
    Interval intervals[] = {{1,3}, {2,6}, {8,10}, {15,18}};
    int n = sizeof(intervals) / sizeof(intervals[0]);
    Interval merged[n];

    int mergedCount = mergeIntervals(intervals, n, merged);

    printf("Merged intervals:\n");
    printIntervals(merged, mergedCount);

    return 0;
}

qsort(intervals, n, sizeof(Interval), compare);

Sort intervals by start time to prepare for merging

merged[0] = intervals[0];

Initialize merged list with first interval

if (intervals[i].start <= merged[idx].end) {

Check if current interval overlaps with last merged interval

if (intervals[i].end > merged[idx].end) { merged[idx].end = intervals[i].end; }

Extend the end of merged interval if current interval ends later

else { idx++; merged[idx] = intervals[i]; }

No overlap, add current interval as new merged interval

Output

Merged intervals:

[1,6] -> [8,10] -> [15,18] -> null

Complexity Analysis

Time: O(n log n) because we sort all intervals first, then merge in one pass O(n)

Space: O(n) for the output array to store merged intervals

vs Alternative: Naive approach compares every pair leading to O(n^2), sorting reduces it to O(n log n)

Edge Cases

empty intervals array

returns zero merged intervals without error

DSA C

if (n == 0) return 0;

all intervals non-overlapping

merged intervals equal original intervals

DSA C

else { idx++; merged[idx] = intervals[i]; }

all intervals fully overlapping

merged intervals is a single interval covering all

DSA C

if (intervals[i].start <= merged[idx].end) { if (intervals[i].end > merged[idx].end) { merged[idx].end = intervals[i].end; } }

When to Use This Pattern

When you see a list of intervals and need to combine overlapping ones, use the merge intervals pattern by sorting and merging neighbors.

Common Mistakes

Mistake: Not sorting intervals before merging, causing incorrect merges
Fix: Always sort intervals by start time before merging

Mistake: Not updating the end of merged interval when current interval extends beyond it
Fix: Update merged interval end to max of current and merged ends

Summary

It merges overlapping intervals into combined intervals without overlaps.

Use it when you need to simplify or combine ranges that may overlap.

Sort intervals by start time first, then merge neighbors by comparing ends.