Update 1235-maximum-profit-in-job-scheduling.kt

a93a · web-flow · commit 2e1aec00fd31 · 2024-01-29T16:45:30.000+01:00
diff --git a/kotlin/1235-maximum-profit-in-job-scheduling.kt b/kotlin/1235-maximum-profit-in-job-scheduling.kt
@@ -1,3 +1,4 @@
+// Recursion + Memoization
 class Solution {
     fun jobScheduling(s: IntArray, e: IntArray, p: IntArray): Int {
         val dp = IntArray (s.size) { -1 }
@@ -35,3 +36,72 @@ class Solution {
         return dfs(0)
     }
 }
+
+// Top-down DP
+class Solution {
+    fun jobScheduling(sT: IntArray, eT: IntArray, p: IntArray): Int {
+        val jobs = sT.indices
+            .map { intArrayOf(sT[it], eT[it], p[it]) }
+            .sortedWith(compareBy({ it[0] }, { it[1] }))
+
+        val n = jobs.size
+        val last = n - 1
+        val dp = IntArray (n)
+
+        for (i in last downTo 0) {
+            var j = -1
+            var l = i
+            var r = last
+            while (l <= r) {
+                val m = (l + r) / 2
+                if (jobs[m][0] >= jobs[i][1]) {
+                    j = m
+                    r = m - 1
+                } else {
+                    l = m + 1
+                }
+            }
+
+            dp[i] = maxOf(
+                jobs[i][2] + (if (j != -1) dp[j] else 0),
+                if (i + 1 < n) dp[i + 1] else 0
+            )
+        }
+
+        return dp[0]
+    }
+}
+
+//Bottom-up DP 
+class Solution {
+    fun jobScheduling(sT: IntArray, eT: IntArray, p: IntArray): Int {
+        val jobs = sT.indices
+            .map { intArrayOf(sT[it], eT[it], p[it]) }
+            .sortedWith(compareBy({ it[1] }, { it[0] }))
+
+        val n = jobs.size
+        val dp = IntArray (n)
+
+        for (i in 0 until n) {
+            var l = 0
+            var r = i - 1
+            var j = -1
+            while (l <= r) {
+                val m = (l + r) / 2
+                if (jobs[m][1] <= jobs[i][0]) {
+                    j = m
+                    l = m + 1
+                } else {
+                    r = m - 1
+                }
+            }
+
+            dp[i] = maxOf(
+                jobs[i][2] + if (j >= 0) dp[j] else 0,
+                if (i - 1 >= 0) dp[i - 1] else 0
+            )
+        }
+
+        return dp[n - 1]
+    }
+}
