Day 27 - Perfect Squares

Author: suriyakrishna

src/main/scala/com/kishan/scala/leetcode/juneChallenges/PerfectSquares.scala

@@ -0,0 +1,64 @@
+package com.kishan.scala.leetcode.juneChallenges
+
+/*
+* Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
+* */
+
+
+/*
+* Input: n = 12
+  Output: 3
+  Explanation: 12 = 4 + 4 + 4.
+*
+* Input: n = 13
+  Output: 2
+  Explanation: 13 = 4 + 9.
+*
+* */
+
+/*
+* Approach: Dynamic Programming
+*
+* Assume we want to find the numSquares for 13. Below will be iterations.
+* Intially for each value the value it self will be the possible value. i.e for 4 = 1^2 + 1^2 + 1^2 + 1^2
+*
+* dp = [0,0,0,0,0,0,0,0,0,0,0,0,0,0]
+* dp(0) -> 0
+* dp(1) -> 1 and inner loop dp(1) -> min(dp(1), 1 + dp(1 - 1*1)) -> 1
+* dp(2) -> 2 and inner loop dp(2) -> min(dp(2), 1 + dp(2 - 1*1)) -> 2
+* dp(3) -> 3 and inner loop dp(3) -> min(dp(3), 1 + dp(3 - 1*1)) -> 3
+* dp(4) -> 4 and inner loop dp(4) -> min(dp(4), 1 + dp(4 - 1*1)) -> 4, dp(4) -> min(dp(4), 1 + dp(4 - 2*2)) -> 1
+* dp(5) -> 5 and inner loop dp(5) -> min(dp(5), 1 + dp(5 - 1*1)) -> 2, dp(5) -> min(dp(5), 1 + dp(5 - 2*2)) -> 2
+* dp(6) -> 6 and inner loop dp(6) -> min(dp(6), 1 + dp(6 - 1*1)) -> 3, dp(6) -> min(dp(6), 1 + dp(6 - 2*2)) -> 3
+* dp(7) -> 7 and inner loop dp(7) -> min(dp(7), 1 + dp(7 - 1*1)) -> 4, dp(7) -> min(dp(7), 1 + dp(7 - 2*2)) -> 4
+* dp(8) -> 8 and inner loop dp(8) -> min(dp(8), 1 + dp(8 - 1*1)) -> 5, dp(8) -> min(dp(8), 1 + dp(8 - 2*2)) -> 2
+* dp(9) -> 9 and inner loop dp(9) -> min(dp(9), 1 + dp(9 - 1*1)) -> 3, dp(9) -> min(dp(9), 1 + dp(9 - 2*2)) -> 3, , dp(9) -> min(dp(9), 1 + dp(9 - 3*3)) -> 1
+* dp(10) -> 10 and inner loop dp(10) -> min(dp(10), 1 + dp(10 - 1*1)) -> 2, dp(10) -> min(dp(10), 1 + dp(10 - 2*2)) -> 2, , dp(10) -> min(dp(10), 1 + dp(10 - 3*3)) -> 2
+* dp(11) -> 11 and inner loop dp(11) -> min(dp(11), 1 + dp(11 - 1*1)) -> 3, dp(11) -> min(dp(11), 1 + dp(11 - 2*2)) -> 3, , dp(11) -> min(dp(11), 1 + dp(11 - 3*3)) -> 3
+* dp(12) -> 12 and inner loop dp(12) -> min(dp(12), 1 + dp(12 - 1*1)) -> 4, dp(12) -> min(dp(12), 1 + dp(12 - 2*2)) -> 3, , dp(12) -> min(dp(12), 1 + dp(12 - 3*3)) -> 3
+* dp(13) -> 13 and inner loop dp(13) -> min(dp(13), 1 + dp(13 - 1*1)) -> 4, dp(13) -> min(dp(13), 1 + dp(13 - 2*2)) -> 2, , dp(12) -> min(dp(12), 1 + dp(13 - 3*3)) -> 2
+*
+*
+* Finally, we will return the n'th value of the dp which will be the result
+* */
+
+
+object PerfectSquares {
+  def main(args: Array[String]): Unit = {
+    val result = numSquares(13)
+    println(s"Result: ${result}, ${result == 3}")
+  }
+
+  def numSquares(n: Int): Int = {
+    var dp: Array[Int] = Array.ofDim[Int](n + 1)
+    for (i <- dp.indices) {
+      dp(i) = i
+      var j = 1
+      while (j * j <= i) {
+        dp(i) = math.min(dp(i), 1 + dp(i - j * j))
+        j += 1
+      }
+    }
+    return dp(n)
+  }
+}