LeetCode //C - 757. Set Intersection Size At Least Two

757. Set Intersection Size At Least Two

You are given a 2D integer array intervals where i n t e r v a l s i = s t a r t i , e n d i intervalsi = start_i, end_i intervalsi=starti,endi represents all the integers from s t a r t i start_i starti to e n d i end_i endi inclusively.

A containing set is an array nums where each interval from intervals has at least two integers in nums.

• For example, if intervals = \[1,3, 3,7, 8,9], then 1,2,4,7,8,9 and 2,3,4,8,9 are containing sets.

Return the minimum possible size of a containing set.

Example 1:

Input: intervals = \[1,3,3,7,8,9]
Output: 5
Explanation: let nums = 2, 3, 4, 8, 9.

It can be shown that there cannot be any containing array of size 4.

Example 2:

Input: intervals = \[1,3,1,4,2,5,3,5]
Output: 3
Explanation: let nums = 2, 3, 4.

It can be shown that there cannot be any containing array of size 2.

Example 3:

Input: intervals = \[1,2,2,3,2,4,4,5]
Output: 5
Explanation: let nums = 1, 2, 3, 4, 5.

It can be shown that there cannot be any containing array of size 4.

Constraints:

• 1 <= intervals.length <= 3000
• intervalsi.length == 2
• 0 < = s t a r t i < e n d i < = 10 8 0 <= start_i < end_i <= 10^8 0<=starti<endi<=108

From: LeetCode

Link: 757. Set Intersection Size At Least Two

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Solution:

Ideas:

• Sort intervals by end point (ascending), then by start point (descending) when end points are equal

• Track state using two variables: point1 and point2 (last two points added to our set)

• Process each interval and check how many of our current points it contains:

  • Contains 0 points (start > point2): Add 2 points at positions end-1 and end
  • Contains 1 point (start > point1): Add 1 point at position end
  • Contains 2+ points: Add 0 points

Code:

// Comparator function to sort intervals by end point
// If end points are equal, sort by start point in descending order
int compare(const void* a, const void* b) {
    int* intervalA = *(int**)a;
    int* intervalB = *(int**)b;
    if (intervalA[1] != intervalB[1]) {
        return intervalA[1] - intervalB[1];  // Sort by end point ascending
    }
    return intervalB[0] - intervalA[0];  // If end points equal, sort by start point descending
}

int intersectionSizeTwo(int** intervals, int intervalsSize, int* intervalsColSize) {
    if (intervalsSize == 0) return 0;
    
    // Sort intervals by end point
    qsort(intervals, intervalsSize, sizeof(int*), compare);
    
    int count = 0;
    int point1 = INT_MIN;  // Second to last point added
    int point2 = INT_MIN;  // Last point added
    
    for (int i = 0; i < intervalsSize; i++) {
        int start = intervals[i][0];
        int end = intervals[i][1];
        
        if (start > point2) {
            // Current interval doesn't contain any of our points
            // Add the two rightmost points of this interval
            count += 2;
            point1 = end - 1;
            point2 = end;
        } else if (start > point1) {
            // Current interval contains only point2
            // Add one more point (the rightmost one)
            count += 1;
            point1 = point2;
            point2 = end;
        }
        // If start <= point1, then the interval contains both point1 and point2
        // No need to add any points
    }
    
    return count;
}