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Day ??? - Graph Search Algorithm III
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import java.util.*; | ||
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public class GraphSearchAlgorithmIII { | ||
/** | ||
* Largest Product Of Length | ||
* <p> | ||
* Given a dictionary containing many words, find the largest product of two | ||
* words’ lengths, such that the two words do not share any common characters. | ||
*/ | ||
public int largestProduct(String[] dict) { | ||
Arrays.sort(dict, new Comparator<String>() { | ||
@Override | ||
public int compare(String a, String b) { | ||
return b.length() - a.length(); | ||
} | ||
}); | ||
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int[] bitmap = buildBitmap(dict); | ||
int n = dict.length; | ||
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PriorityQueue<Pair> pq = new PriorityQueue<>(); | ||
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pq.offer(new Pair(0, 0, dict, bitmap)); | ||
boolean[][] visited = new boolean[n][n]; | ||
visited[0][0] = true; | ||
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while (!pq.isEmpty()) { | ||
Pair cur = pq.poll(); | ||
if (!cur.hasCommon()) { | ||
return cur.prod(); | ||
} | ||
int i = cur.i; | ||
int j = cur.j; | ||
if (i < n - 1 && !visited[i + 1][j]) { | ||
visited[i + 1][j] = true; | ||
pq.offer(new Pair(i + 1, j, dict, bitmap)); | ||
} | ||
if (j < n - 1 && !visited[i][j + 1]) { | ||
visited[i][j + 1] = true; | ||
pq.offer(new Pair(i, j + 1, dict, bitmap)); | ||
} | ||
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} | ||
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return 0; | ||
} | ||
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private class Pair implements Comparable<Pair> { | ||
public int i; | ||
public int j; | ||
private String[] dict; | ||
private int[] bitmap; | ||
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public Pair(int i, int j, String[] dict, int[] bitmap) { | ||
this.i = i; | ||
this.j = j; | ||
this.dict = dict; | ||
this.bitmap = bitmap; | ||
} | ||
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@Override | ||
public int compareTo(Pair other) { | ||
return other.prod() - this.prod(); | ||
} | ||
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public int prod() { | ||
return dict[i].length() * dict[j].length(); | ||
} | ||
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public boolean hasCommon() { | ||
return (bitmap[i] & bitmap[j]) != 0; | ||
} | ||
} | ||
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private int[] buildBitmap(String[] dict) { | ||
int[] bitmap = new int[dict.length]; | ||
for (int i = 0; i < dict.length; i++) { | ||
String cur = dict[i]; | ||
int bits = 0; | ||
for (int j = 0; j < cur.length(); j++) { | ||
bits |= (1 << (cur.charAt(j) - 'a')); | ||
} | ||
bitmap[i] = bits; | ||
} | ||
return bitmap; | ||
} | ||
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/** | ||
* Kth Smallest With Only 3, 5, 7 As Factors | ||
* <p> | ||
* Find the Kth smallest number s such that s = 3 ^ x * 5 ^ y * 7 ^ z, x > 0 and | ||
* y > 0 and z > 0, x, y, z are all integers. | ||
*/ | ||
public long kth(int k) { | ||
if (k == 1) { | ||
return 105L; | ||
} | ||
PriorityQueue<Long> pq = new PriorityQueue<>(); | ||
Set<Long> set = new HashSet<>(); | ||
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pq.offer(105L); | ||
while (--k > 0) { | ||
long cur = pq.poll(); | ||
if (!set.contains(cur * 3)) { | ||
pq.offer(cur * 3); | ||
set.add(cur * 3); | ||
} | ||
if (!set.contains(cur * 5)) { | ||
pq.offer(cur * 5); | ||
set.add(cur * 5); | ||
} | ||
if (!set.contains(cur * 7)) { | ||
pq.offer(cur * 7); | ||
set.add(cur * 7); | ||
} | ||
} | ||
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return pq.peek(); | ||
} | ||
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/** | ||
* Kth Closest Point To <0,0,0> | ||
* <p> | ||
* Given three arrays sorted in ascending order. Pull one number from each array | ||
* to form a coordinate <x,y,z> in a 3D space. Find the coordinates of the | ||
* points that is k-th closest to <0,0,0>. | ||
*/ | ||
public List<Integer> closest(int[] a, int[] b, int[] c, int k) { | ||
int al = a.length; | ||
int bl = b.length; | ||
int cl = c.length; | ||
boolean[][][] visited = new boolean[al][bl][cl]; | ||
PriorityQueue<Coord> pq = new PriorityQueue<>(); | ||
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return null; | ||
} | ||
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private class Coord implements Comparable<Coord> { | ||
@Override | ||
public int compareTo(Coord other) { | ||
return 0; | ||
} | ||
} | ||
} |