11package com .interview .dynamic ;
22
3+ import java .util .*;
4+ 35/**
6+ * Date 10/19/2016
7+ * @author Tushar Roy
8+ *
9+ * Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
10+ * For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
11+ *
12+ * Solution 1 - Using DP similar to coin change problem with infinite supply
13+ * Solution 2 - Using a BFS. Put all perfect squares in queue. Then considering each as a node try adding
14+ * another perfect square and see if we can get n. Keep doing this in BFS fashion till you hit the number.
15+ *
416 * https://leetcode.com/problems/perfect-squares/
517 */
618public class MinimumNumberOfPerfectSquares {
7- public int numSquares (int n ) {
19+ public int numSquaresUsingDP (int n ) {
820 int count = (int )Math .ceil (Math .sqrt (n ));
921
1022 int [] T = new int [n + 1 ];
@@ -22,4 +34,44 @@ public int numSquares(int n) {
2234 }
2335 return T [n ];
2436 }
37+ 38+ public int numSquaresUsingBFS (int n ) {
39+ List <Integer > perfectSquares = new ArrayList <>();
40+ int i = 1 ;
41+ Queue <Integer > queue = new LinkedList <>();
42+ Set <Integer > visited = new HashSet <>();
43+ while (true ) {
44+ int square = i * i ;
45+ if (square == n ) {
46+ return 1 ;
47+ }
48+ if (square > n ) {
49+ break ;
50+ }
51+ perfectSquares .add (square );
52+ queue .offer (square );
53+ visited .add (square );
54+ i ++;
55+ }
56+ int distance = 1 ;
57+ while (!queue .isEmpty ()) {
58+ int size = queue .size ();
59+ distance ++;
60+ for (int j = 0 ; j < size ; j ++) {
61+ int node = queue .poll ();
62+ for (int square : perfectSquares ) {
63+ int sum = node + square ;
64+ if (sum == n ) {
65+ return distance ;
66+ } else if (!visited .contains (sum )) {
67+ visited .add (sum );
68+ queue .add (sum );
69+ } else if (sum > n ) {
70+ break ;
71+ }
72+ }
73+ }
74+ }
75+ return distance ;
76+ }
2577}
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