update first_missing_positive

Author: kumailn

Kotlin/First_Missing_Positive.kt

@@ -0,0 +1,31 @@
+// Question: Given a list of ints, find the smallest missing positive number
+// Solution: Sort the list as you go through by placing every term on its index + 1th place.
+// Difficulty: Hard 
+// Time Complexity: O(n)
+// Space Complexity: O(1)
+
+fun firstMissingPositive(nums: IntArray): Int {
+    var left = 0;
+    var right = nums.size;
+    
+    while (left < right) {
+        // If the current number is equal to its index + 1, skip over as its already sorted
+        // for example if the number 5 is in the 4th index/ 5th place
+        if (nums[left] == left + 1) left++;
+
+        // Otherwise, if  number not in the correct place
+        // In othe words, if a number n is within our range and the number in the nth position isnt equal to this number, swap the two
+        else if (nums[left] > 0 && nums[left] <= nums.size && nums[left] != nums[nums[left]-1]) {
+            nums[nums[left]-1] = nums[left].also{ nums[left] = nums[nums[left]-1] };
+        } 
+        
+        // Case if the number is a duplicate, or out of our range (1 to right). So place it at the end 
+        else nums[left] = nums[--right].also{ nums[right] = nums[left] };   
+    }
+    return right + 1;
+}
+
+fun main(args: Array<String>) { 
+    var arr: IntArray = intArrayOf(7,8,9,11,12);
+    print(firstMissingPositive(arr));
+}

Python/First_Missing_Positive.py

@@ -1,27 +1,30 @@
 #Question: Given a list of ints, find the smallest missing positive number
 #Solution: Sort the list as you go through by placing every term on its index + 1th place.
 #Difficulty: Hard 
+#Time Complexity: O(n)
+#Space Complexity: O(1)
 
 from typing import List
 
 def firstMissingPositive(nums: List[int]) -> int:
-        i, j = 0, len(nums)
-        while i < j:
+    i, j = 0, len(nums)
+    while i < j:
 
-            #If the current number is equal to its index + 1, skip over as its already sorted
-            #ie if the number 5 is in the 4th index/ 5th place
-            if nums[i] == i+1: i += 1
+        #If the current number is equal to its index + 1, skip over as its already sorted
+        #for example if the number 5 is in the 4th index/ 5th place
+        if nums[i] == i+1: i += 1
 
-            #Regular case, ie a number not in the correct place
-            elif nums[i] > 0 and nums[i] <= j and nums[i] != nums[nums[i]-1]: 
-                nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
+        #Regular case, ie a number not in the correct place
+        #In othe words, if a number n is within our range and the number in the nth position isnt equal to this number, swap the two
+        elif nums[i] > 0 and nums[i] <= j and nums[nums[i]-1] != nums[i]: 
+            nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
 
-            #Case if the number is a duplicate, or out of our range (1 to j). So place it at the end 
-            else: 
-                j -= 1
-                nums[j], nums[i] = nums[i], nums[j]
+        #Case if the number is a duplicate, or out of our range (1 to j). So place it at the end 
+        else: 
+            j -= 1
+            nums[j], nums[i] = nums[i], nums[j]
 
-        return j + 1
+    return j + 1
 
 def main():
     print(firstMissingPositive([1, 5, 4, 3, 2, 7, 0]))