-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathSegmentTree.cpp
More file actions
632 lines (497 loc) · 15.5 KB
/
SegmentTree.cpp
File metadata and controls
632 lines (497 loc) · 15.5 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
#include <bits/stdc++.h>
using namespace std;
// Segment Tree, Basic Implementations=================================================
/*
https://cp-algorithms.com/data_structures/segment_tree.html#toc-tgt-6
Memory efficient implementation
Most people use the implementation from the previous section.
If you look at the array t you can see that it follows the numbering of the tree nodes
in the order of a BFS traversal (level-order traversal).
Using this traversal the children of vertex v are 2v and 2v+1 respectively.
However if n is not a power of two, this method will skip some indices
and leave some parts of the array t unused. The memory consumption is limited by 4n,
even though a Segment Tree of an array of n elements requires only 2n−1 vertices.
However it can be reduced.
We renumber the vertices of the tree in the order of an Euler tour traversal (pre-order traversal),
and we write all these vertices next to each other.
Lets look at a vertex at index v,
and let him be responsible for the segment [l,r], and let mid=l+r2.
It is obvious that the left child will have the index v+1.
The left child is responsible for the segment [l,mid],
i.e. in total there will be 2∗(mid−l+1)−1 vertices in the left child's subtree.
Thus we can compute the index of the right child of v.
The index will be v+2∗(mid−l+1).
By this numbering we achieve a reduction of the necessary memory to 2n.
*/
// Range Maximum Query with Node Update
// (https://www.pepcoding.com/resources/data-structures-and-algorithms-in-java-interview-prep/segment-tree/max-in-a-interval-range-query-point-update-official/ojquestion)
/*
Approach:
Update: O(logn)
Query: O(logn)
*/
class SegmentTree
{
private:
vector<int> tree;
vector<int> arr;
void buildTree(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
tree[nodeIdx] = arr[lo];
return;
}
int mid = lo + (hi - lo) / 2;
buildTree(nodeIdx * 2, lo, mid); // left child
buildTree(nodeIdx * 2 + 1, mid + 1, hi); // right child
//current node's max is max of left and right child
tree[nodeIdx] = max(tree[nodeIdx * 2], tree[nodeIdx * 2 + 1]);
}
int query(int nodeIdx, int lo, int hi, int left, int right)
{
// current interval is outside target interval
if (hi < left || lo > right)
return INT_MIN;
// leaf node OR current interval is completely inside target interval
if (lo == hi || (lo >= left && hi <= right))
return tree[nodeIdx];
int mid = lo + (hi - lo) / 2;
// get the left and right max
int leftMax = query(nodeIdx * 2, lo, mid, left, right);
int rightMax = query(nodeIdx * 2 + 1, mid + 1, hi, left, right);
// return the max of both children
return max(leftMax, rightMax);
}
void update(int nodeIdx, int lo, int hi, int pos, int val)
{
// base case
if (lo == hi)
{
tree[nodeIdx] = val;
arr[pos] = val;
return;
}
int mid = lo + (hi - lo) / 2;
if (pos > mid) // go to right child
update(nodeIdx * 2 + 1, mid + 1, hi, pos, val);
else // go to left child
update(nodeIdx * 2, lo, mid, pos, val);
// update current node value
tree[nodeIdx] = max(tree[nodeIdx * 2], tree[nodeIdx * 2 + 1]);
}
public:
SegmentTree(vector<int> &arr)
{
this->arr = arr;
tree.resize(4 * arr.size());
buildTree(1, 0, arr.size() - 1);
}
void update(int pos, int val)
{
update(1, 0, arr.size() - 1, pos, val);
}
int query(int l, int r)
{
return query(1, 0, arr.size() - 1, l, r);
}
};
// 307. Range Sum Query - Mutable
// Approach 1: O(4*n) Segment Tree Size
class NumArray
{
private:
vector<int> tree;
vector<int> nums;
void buildTree(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
tree[nodeIdx] = nums[lo];
return;
}
int mid = lo + (hi - lo) / 2;
buildTree(2 * nodeIdx, lo, mid);
buildTree(2 * nodeIdx + 1, mid + 1, hi);
tree[nodeIdx] = tree[2 * nodeIdx] + tree[2 * nodeIdx + 1];
}
void update(int nodeIdx, int lo, int hi, int index, int val)
{
if (lo == hi)
{
tree[nodeIdx] = val;
nums[index] = val;
return;
}
int mid = lo + (hi - lo) / 2;
if (index > mid)
update(2 * nodeIdx + 1, mid + 1, hi, index, val);
else
update(2 * nodeIdx, lo, mid, index, val);
tree[nodeIdx] = tree[2 * nodeIdx] + tree[2 * nodeIdx + 1];
}
int sumRange(int nodeIdx, int lo, int hi, int left, int right)
{
if (lo > right || hi < left)
return 0;
if (lo == hi || lo >= left && hi <= right)
return tree[nodeIdx];
int mid = lo + (hi - lo) / 2;
int leftSum = sumRange(2 * nodeIdx, lo, mid, left, right);
int rightSum = sumRange(2 * nodeIdx + 1, mid + 1, hi, left, right);
return leftSum + rightSum;
}
public:
NumArray(vector<int> &nums)
{
this->nums = nums;
this->tree.resize(4 * nums.size());
buildTree(1, 0, nums.size() - 1);
}
void update(int index, int val)
{
update(1, 0, this->nums.size() - 1, index, val);
}
int sumRange(int left, int right)
{
return sumRange(1, 0, this->nums.size() - 1, left, right);
}
};
// Approach 2: Same approach with O(2*n) Segment Tree Size
/*
Instead of storing left child at 2*node, right at 2*node + 1
Store in preorder traversal order
So, left = node + 1
right = node + 2*(mid - lo + 1)
Rest everything remains the exact same
*/
class NumArray
{
private:
vector<int> tree;
vector<int> nums;
void buildTree(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
tree[nodeIdx] = nums[lo];
return;
}
int mid = lo + (hi - lo) / 2;
buildTree(nodeIdx + 1, lo, mid);
buildTree(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi);
tree[nodeIdx] = tree[nodeIdx + 1] + tree[nodeIdx + 2 * (mid - lo + 1)];
}
void update(int nodeIdx, int lo, int hi, int index, int val)
{
if (lo == hi)
{
tree[nodeIdx] = val;
nums[index] = val;
return;
}
int mid = lo + (hi - lo) / 2;
if (index > mid)
update(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, index, val);
else
update(nodeIdx + 1, lo, mid, index, val);
tree[nodeIdx] = tree[nodeIdx + 1] + tree[nodeIdx + 2 * (mid - lo + 1)];
}
int sumRange(int nodeIdx, int lo, int hi, int left, int right)
{
if (lo > right || hi < left)
return 0;
if (lo == hi || lo >= left && hi <= right)
return tree[nodeIdx];
int mid = lo + (hi - lo) / 2;
int leftSum = sumRange(nodeIdx + 1, lo, mid, left, right);
int rightSum = sumRange(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, left, right);
return leftSum + rightSum;
}
public:
NumArray(vector<int> &nums)
{
this->nums = nums;
this->tree.resize(2 * nums.size());
buildTree(1, 0, nums.size() - 1);
}
void update(int index, int val)
{
update(1, 0, this->nums.size() - 1, index, val);
}
int sumRange(int left, int right)
{
return sumRange(1, 0, this->nums.size() - 1, left, right);
}
};
// Lazy Propagation===============================================================
// What's At Idx - Point Query Range Update
// (https://www.pepcoding.com/resources/data-structures-and-algorithms-in-java-interview-prep/segment-tree/whats-at-idx-point-query-range-update-official/ojquestion)
class SegmentTree
{
private:
vector<int> lazy, arr;
void build(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
lazy[nodeIdx] = arr[lo];
return;
}
int mid = lo + (hi - lo) / 2;
build(nodeIdx + 1, lo, mid);
build(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi);
lazy[nodeIdx] = 0;
}
void update(int nodeIdx, int lo, int hi, int l, int r, int val)
{
if (hi < l || lo > r)
return;
if (lo == hi || l <= lo && hi <= r)
{
lazy[nodeIdx] += val;
return;
}
int mid = lo + (hi - lo) / 2;
update(nodeIdx + 1, lo, mid, l, r, val);
update(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, l, r, val);
}
void propagate(int nodeIdx, int lo, int hi)
{
// if leaf node, then return
if (lo == hi)
return;
int mid = lo + (hi - lo) / 2;
// propagate to left child
lazy[nodeIdx + 1] += lazy[nodeIdx];
// propagate to right child
lazy[nodeIdx + 2 * (mid - lo + 1)] += lazy[nodeIdx];
// reset current node to 0
lazy[nodeIdx] = 0;
}
int query(int nodeIdx, int lo, int hi, int idx)
{
propagate(nodeIdx, lo, hi);
if (lo == hi)
return lazy[nodeIdx];
int mid = lo + (hi - lo) / 2;
if (idx > mid)
return query(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, idx);
else
return query(nodeIdx + 1, lo, mid, idx);
}
public:
SegmentTree(vector<int> &arr)
{
this->arr = arr;
lazy.resize(2 * arr.size());
build(1, 0, arr.size() - 1);
}
void update(int l, int r, int val)
{
update(1, 0, arr.size() - 1, l, r, val);
}
int query(int idx)
{
return query(1, 0, arr.size() - 1, idx);
}
};
// Sum Of Range - Range Query Range Update
// (https://www.pepcoding.com/resources/data-structures-and-algorithms-in-java-interview-prep/segment-tree/sum-of-range-range-query-range-update-official/ojquestion)
class SegmentTree
{
private:
vector<int> lazy, arr, tree;
void build(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
tree[nodeIdx] = arr[lo];
return;
}
int mid = lo + (hi - lo) / 2;
build(nodeIdx + 1, lo, mid);
build(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi);
lazy[nodeIdx] = 0;
tree[nodeIdx] = tree[nodeIdx + 1] + tree[nodeIdx + 2 * (mid - lo + 1)];
}
void update(int nodeIdx, int lo, int hi, int l, int r, int val)
{
propagate(nodeIdx, lo, hi);
if (hi < l || lo > r)
return;
if (lo == hi)
tree[nodeIdx] += val;
else if (l <= lo && hi <= r)
{
lazy[nodeIdx] += val;
propagate(nodeIdx, lo, hi);
}
else
{
int mid = lo + (hi - lo) / 2;
update(nodeIdx + 1, lo, mid, l, r, val);
update(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, l, r, val);
tree[nodeIdx] = tree[nodeIdx + 1] + tree[nodeIdx + 2 * (mid - lo + 1)];
}
}
void propagate(int nodeIdx, int lo, int hi)
{
// if leaf node, then return
if (lo == hi)
{
tree[nodeIdx] += lazy[nodeIdx];
lazy[nodeIdx] = 0;
}
int mid = lo + (hi - lo) / 2;
// update the sum in tree
tree[nodeIdx] += lazy[nodeIdx] * (hi - lo + 1);
// propagate to left child
lazy[nodeIdx + 1] += lazy[nodeIdx];
// propagate to right child
lazy[nodeIdx + 2 * (mid - lo + 1)] += lazy[nodeIdx];
// reset current node to 0
lazy[nodeIdx] = 0;
}
int query(int nodeIdx, int lo, int hi, int l, int r)
{
propagate(nodeIdx, lo, hi);
if (hi < l || lo > r)
return 0;
if (lo == hi || l <= lo && hi <= r)
return tree[nodeIdx];
int mid = lo + (hi - lo) / 2;
int rightSum = query(nodeIdx + 2 * (mid - lo + 1), mid + 1, hi, l, r);
int leftSum = query(nodeIdx + 1, lo, mid, l, r);
return leftSum + rightSum;
}
public:
SegmentTree(vector<int> &arr)
{
this->arr = arr;
lazy.resize(2 * arr.size());
tree.resize(2 * arr.size());
build(1, 0, arr.size() - 1);
}
void update(int l, int r, int val)
{
update(1, 0, arr.size() - 1, l, r, val);
}
int query(int l, int r)
{
return query(1, 0, arr.size() - 1, l, r);
}
};
// Maximum Sum (https://www.spoj.com/problems/KGSS/)
/*
Approach: O(logn)
Store the {maxSum, maxVal} in each node
where maxSum = max val + 2nd max val
So, for build each node will have the value
maxSum = max(max sum of left child, max sum of right child, max val of left + max val of right)
maxVal = max(max val of left child, max val of right child)
And same for query, we will get the {maxSum, maxVal} from both children and
return the max of both
*/
class SegmentTree
{
private:
vector<pair<int, int>> tree;
vector<int> arr;
void updateCurrentNode(int nodeIdx)
{
// every node stores {max sum of 2 values, max value}
pair<int, int> left = tree[2 * nodeIdx];
pair<int, int> right = tree[2 * nodeIdx + 1];
int maxSum = max({left.first, right.first, left.second + right.second});
int maxVal = max(left.second, right.second);
tree[nodeIdx] = {maxSum, maxVal};
}
void build(int nodeIdx, int lo, int hi)
{
if (lo == hi)
{
tree[nodeIdx] = {arr[lo], arr[lo]};
return;
}
int mid = lo + (hi - lo) / 2;
build(2 * nodeIdx, lo, mid);
build(2 * nodeIdx + 1, mid + 1, hi);
updateCurrentNode(nodeIdx);
}
void update(int nodeIdx, int lo, int hi, int idx, int val)
{
if (lo == hi)
{
tree[nodeIdx] = {val, val};
arr[idx] = val;
return;
}
int mid = lo + (hi - lo) / 2;
if (idx > mid)
update(2 * nodeIdx + 1, mid + 1, hi, idx, val);
else
update(2 * nodeIdx, lo, mid, idx, val);
updateCurrentNode(nodeIdx);
}
pair<int, int> query(int nodeIdx, int lo, int hi, int l, int r)
{
if (hi < l || lo > r)
return {INT_MIN, INT_MIN};
if (lo == hi || l <= lo && hi <= r)
return tree[nodeIdx];
int mid = lo + (hi - lo) / 2;
pair<int, int> leftMax = query(2 * nodeIdx, lo, mid, l, r);
pair<int, int> rightMax = query(2 * nodeIdx + 1, mid + 1, hi, l, r);
int maxSum = max({leftMax.first, rightMax.first, leftMax.second + rightMax.second});
int maxVal = max(leftMax.second, rightMax.second);
return {maxSum, maxVal};
}
public:
SegmentTree(vector<int> &arr)
{
this->arr = arr;
this->tree.resize(4 * arr.size());
build(1, 0, arr.size() - 1);
}
void update(int idx, int val)
{
update(1, 0, arr.size() - 1, idx, val);
}
int query(int l, int r)
{
return query(1, 0, arr.size() - 1, l, r).first;
}
};
main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// input
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++)
cin >> arr[i];
SegmentTree *treeObj = new SegmentTree(arr);
int q;
cin >> q;
while (q--)
{
char x;
cin >> x;
if (x == 'U')
{
int pos, val;
cin >> pos >> val;
treeObj->update(pos, val);
}
else
{
int l, r;
cin >> l >> r;
cout << treeObj->query(l, r) << endl;
}
}
return 0;
}