Improved task 2948

Author: javadev

src/main/kotlin/g2901_3000/s2948_make_lexicographically_smallest_array_by_swapping_elements/Solution.kt

@@ -1,42 +1,39 @@
 package g2901_3000.s2948_make_lexicographically_smallest_array_by_swapping_elements
 
-// #Medium #Array #Sorting #Union_Find #2023_12_31_Time_2141_ms_(6.45%)_Space_69.2_MB_(74.19%)
+// #Medium #Array #Sorting #Union_Find #2023_12_31_Time_1831_ms_(6.45%)_Space_69.4_MB_(74.19%)
 
 import kotlin.math.abs
 
 class Solution {
     fun lexicographicallySmallestArray(nums: IntArray, limit: Int): IntArray {
         val n = nums.size
-        val nodes = arrayOfNulls<Node>(n)
-        for (i in 0 until n) {
-            nodes[i] = Node(i, nums[i])
-        }
-        nodes.sortWith { a: Node?, b: Node? ->
+        val nodes = Array(n) { i -> Node(i, nums[i]) }
+        nodes.sortWith { a: Node, b: Node ->
             Integer.signum(
-                a!!.value - b!!.value
+                a.value - b.value
             )
         }
         var group = 1
-        nodes[0]!!.group = group
+        nodes[0].group = group
         for (i in 1 until n) {
-            if (abs((nodes[i]!!.value - nodes[i - 1]!!.value).toDouble()) <= limit) {
-                nodes[i]!!.group = group
+            if (abs(nodes[i].value - nodes[i - 1].value) <= limit) {
+                nodes[i].group = group
             } else {
-                nodes[i]!!.group = ++group
+                nodes[i].group = ++group
             }
         }
         val groupBase = IntArray(group + 1)
         for (i in n - 1 downTo 0) {
-            groupBase[nodes[i]!!.group] = i
+            groupBase[nodes[i].group] = i
         }
         val groupIndex = IntArray(n)
         for (node in nodes) {
-            groupIndex[node!!.id] = node.group
+            groupIndex[node.id] = node.group
         }
         val ans = IntArray(n)
         for (i in 0 until n) {
             val index = groupBase[groupIndex[i]]
-            ans[i] = nodes[index]!!.value
+            ans[i] = nodes[index].value
             groupBase[groupIndex[i]]++
         }
         return ans