480. Sliding Window Median
Description
The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.
- For examples, if
arr = [2,3,4]
, the median is3
. - For examples, if
arr = [1,2,3,4]
, the median is(2 + 3) / 2 = 2.5
.
You are given an integer array nums
and an integer k
. There is a sliding window of size k
which is moving from the very left of the array to the very right. You can only see the k
numbers in the window. Each time the sliding window moves right by one position.
Return the median array for each window in the original array. Answers within 10-5
of the actual value will be accepted.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000] Explanation: Window position Median --------------- ----- [1 3 -1] -3 5 3 6 7 1 1 [3 -1 -3] 5 3 6 7 -1 1 3 [-1 -3 5] 3 6 7 -1 1 3 -1 [-3 5 3] 6 7 3 1 3 -1 -3 [5 3 6] 7 5 1 3 -1 -3 5 [3 6 7] 6
Example 2:
Input: nums = [1,2,3,4,2,3,1,4,2], k = 3 Output: [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]
Constraints:
1 <= k <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
Solutions
Solution 1: Dual Priority Queues (Min-Heap and Max-Heap) + Lazy Deletion
We can use two priority queues (min-heap and max-heap) to maintain the elements in the current window. One priority queue stores the smaller half of the elements, and the other priority queue stores the larger half of the elements. This way, the median of the current window is either the average of the top elements of the two heaps or one of the top elements.
We design a class $\textit{MedianFinder}$ to maintain the elements in the current window. This class includes the following methods:
add_num(num)
: Adds $\textit{num}$ to the current window.find_median()
: Returns the median of the elements in the current window.remove_num(num)
: Removes $\textit{num}$ from the current window.prune(pq)
: If the top element of the heap is in the lazy deletion dictionary $\textit{delayed}$, it pops the top element from the heap and decrements its lazy deletion count. If the lazy deletion count of the element becomes zero, it removes the element from the lazy deletion dictionary.rebalance()
: If the number of elements in the smaller half exceeds the number of elements in the larger half by $2$, it moves the top element of the larger half to the smaller half. If the number of elements in the smaller half is less than the number of elements in the larger half, it moves the top element of the larger half to the smaller half.
In the add_num(num)
method, we first consider adding $\textit{num}$ to the smaller half. If $\textit{num}$ is greater than the top element of the larger half, we add $\textit{num}$ to the larger half. Then we call the rebalance()
method to ensure that the size difference between the two priority queues does not exceed $1$.
In the remove_num(num)
method, we increment the lazy deletion count of $\textit{num}$. Then we compare $\textit{num}$ with the top element of the smaller half. If $\textit{num}$ is less than or equal to the top element of the smaller half, we update the size of the smaller half and call the prune()
method to ensure that the top element of the smaller half is not in the lazy deletion dictionary. Otherwise, we update the size of the larger half and call the prune()
method to ensure that the top element of the larger half is not in the lazy deletion dictionary.
In the find_median()
method, if the current window size is odd, we return the top element of the smaller half; otherwise, we return the average of the top elements of the smaller half and the larger half.
In the prune(pq)
method, if the top element of the heap is in the lazy deletion dictionary, it pops the top element from the heap and decrements its lazy deletion count. If the lazy deletion count of the element becomes zero, it removes the element from the lazy deletion dictionary.
In the rebalance()
method, if the number of elements in the smaller half exceeds the number of elements in the larger half by $2$, it moves the top element of the larger half to the smaller half. If the number of elements in the smaller half is less than the number of elements in the larger half, it moves the top element of the larger half to the smaller half.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
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 |
|
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 |
|
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 |
|
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 |
|
Solution 2: Ordered Set
We can use two ordered sets to maintain the elements in the current window. The ordered set $l$ stores the smaller half of the elements in the current window, and the ordered set $r$ stores the larger half of the elements.
We traverse the array $\textit{nums}$. For each element $x$, we add it to the ordered set $r$, then move the smallest element in the ordered set $r$ to the ordered set $l$. If the size of the ordered set $l$ is greater than the size of the ordered set $r$ by more than $1$, we move the largest element in the ordered set $l$ to the ordered set $r$.
If the total number of elements in the current window is $k$ and the size is odd, the maximum value in the ordered set $l$ is the median. If the size of the current window is even, the average of the maximum value in the ordered set $l$ and the minimum value in the ordered set $r$ is the median. Then, we remove the leftmost element of the window and continue traversing the array.
The time complexity is $O(n \log k)$, and the space complexity is $O(k)$. Here, $n$ is the length of the array $\textit{nums}$.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
|
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 |
|
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 |
|
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 |
|
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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 |
|