352. Data Stream as Disjoint Intervals
Given a data stream input of non-negative integers a 1 , a 2 , . . . , a n a_1, a_2, ..., a_n a1,a2,...,an, summarize the numbers seen so far as a list of disjoint intervals.
Implement the SummaryRanges class:
- SummaryRanges() Initializes the object with an empty stream.
- void addNum(int value) Adds the integer value to the stream.
- int\[\]\[\] getIntervals() Returns a summary of the integers in the stream currently as a list of disjoint intervals s t a r t i , e n d i start_i, end_i starti,endi. The answer should be sorted by s t a r t i start_i starti.
Example 1:
Input:
"SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"
\[\], \[1\], \[\], \[3\], \[\], \[7\], \[\], \[2\], \[\], \[6\], \[\]
Output:
null, null, \[\[1, 1\]\], null, \[\[1, 1\], \[3, 3\]\], null, \[\[1, 1\], \[3, 3\], \[7, 7\]\], null, \[\[1, 3\], \[7, 7\]\], null, \[\[1, 3\], \[6, 7\]\]
Explanation
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1); // arr = 1
summaryRanges.getIntervals(); // return \[1, 1]
summaryRanges.addNum(3); // arr = 1, 3
summaryRanges.getIntervals(); // return \[1, 1, 3, 3]
summaryRanges.addNum(7); // arr = 1, 3, 7
summaryRanges.getIntervals(); // return \[1, 1, 3, 3, 7, 7]
summaryRanges.addNum(2); // arr = 1, 2, 3, 7
summaryRanges.getIntervals(); // return \[1, 3, 7, 7]
summaryRanges.addNum(6); // arr = 1, 2, 3, 6, 7
summaryRanges.getIntervals(); // return \[1, 3, 6, 7]
Constraints:
- 0 < = v a l u e < = 1 0 4 0 <= value <= 10^4 0<=value<=104
- At most 3 ∗ 1 0 4 3 * 10^4 3∗104 calls will be made to addNum and getIntervals.
- At most 1 0 2 10^2 102 calls will be made to getIntervals.
From: LeetCode
Link: 352. Data Stream as Disjoint Intervals
Solution:
Ideas:
1. Intervals Representation:
- Intervals are stored in a 2D array intervals, where each interval is represented as a pair start, end.
- For example, the intervals for numbers 1, 3, 7 would be stored as \[1, 1, 3, 3, 7, 7].
2. Efficient Insertion:
- When a new number is added, the code finds the appropriate place to insert it. The intervals are maintained in sorted order, so the number is compared to the existing intervals.
- Depending on whether the new number is adjacent to an existing interval, it either extends or merges the intervals.
3. Merging Logic:
- Merge with the previous interval: If the new number is just after the previous interval (i.e., new number == previous interval end + 1), then the previous interval is extended.
- Merge with the next interval: If the new number is just before the next interval (i.e., new number == next interval start - 1), then the next interval is extended.
- Merge both previous and next intervals: If the new number is adjacent to both the previous and the next intervals, the two intervals are merged into one.
- New Interval: If the new number is not adjacent to any existing intervals, a new interval is created.
4. Dynamic Array Management:
- The intervals array has an initial capacity, and when it fills up, it is dynamically resized to accommodate more intervals. This ensures that the solution can handle up to the maximum number of intervals allowed by the problem constraints.
Code:
c
typedef struct {
int** intervals; // To store the intervals as a 2D array
int size; // The current number of intervals
int capacity; // The allocated capacity of the intervals array
} SummaryRanges;
SummaryRanges* summaryRangesCreate() {
SummaryRanges* obj = (SummaryRanges*)malloc(sizeof(SummaryRanges));
obj->size = 0;
obj->capacity = 10; // Initial capacity
obj->intervals = (int**)malloc(sizeof(int*) * obj->capacity);
for (int i = 0; i < obj->capacity; ++i) {
obj->intervals[i] = (int*)malloc(sizeof(int) * 2); // Each interval has two elements [start, end]
}
return obj;
}
void summaryRangesAddNum(SummaryRanges* obj, int value) {
int i = 0;
// Find the position to insert or merge intervals
while (i < obj->size && obj->intervals[i][1] < value) {
i++;
}
// Check if value is already included in an interval
if (i < obj->size && obj->intervals[i][0] <= value && obj->intervals[i][1] >= value) {
return;
}
// Merge with the previous and next intervals if possible
int mergeWithPrev = (i > 0 && obj->intervals[i - 1][1] + 1 == value);
int mergeWithNext = (i < obj->size && obj->intervals[i][0] - 1 == value);
if (mergeWithPrev && mergeWithNext) {
// Merge both previous and next intervals
obj->intervals[i - 1][1] = obj->intervals[i][1];
// Remove the current interval
for (int j = i; j < obj->size - 1; ++j) {
obj->intervals[j][0] = obj->intervals[j + 1][0];
obj->intervals[j][1] = obj->intervals[j + 1][1];
}
obj->size--;
} else if (mergeWithPrev) {
// Merge with the previous interval
obj->intervals[i - 1][1] = value;
} else if (mergeWithNext) {
// Merge with the next interval
obj->intervals[i][0] = value;
} else {
// Insert a new interval
if (obj->size == obj->capacity) {
obj->capacity *= 2;
obj->intervals = (int**)realloc(obj->intervals, sizeof(int*) * obj->capacity);
for (int j = obj->size; j < obj->capacity; ++j) {
obj->intervals[j] = (int*)malloc(sizeof(int) * 2);
}
}
for (int j = obj->size; j > i; --j) {
obj->intervals[j][0] = obj->intervals[j - 1][0];
obj->intervals[j][1] = obj->intervals[j - 1][1];
}
obj->intervals[i][0] = value;
obj->intervals[i][1] = value;
obj->size++;
}
}
int** summaryRangesGetIntervals(SummaryRanges* obj, int* retSize, int** retColSize) {
*retSize = obj->size;
*retColSize = (int*)malloc(sizeof(int) * obj->size);
for (int i = 0; i < obj->size; ++i) {
(*retColSize)[i] = 2; // Each interval has two columns
}
return obj->intervals;
}
void summaryRangesFree(SummaryRanges* obj) {
for (int i = 0; i < obj->capacity; ++i) {
free(obj->intervals[i]);
}
free(obj->intervals);
free(obj);
}
/**
* Your SummaryRanges struct will be instantiated and called as such:
* SummaryRanges* obj = summaryRangesCreate();
* summaryRangesAddNum(obj, value);
* int** param_2 = summaryRangesGetIntervals(obj, retSize, retColSize);
* summaryRangesFree(obj);
*/