Updated tags for tasks 41-48

Author: javadev

src/main/go/g0001_0100/s0041_first_missing_positive/readme.md

@@ -2,33 +2,33 @@
 
 Hard
 
-Given an unsorted integer array `nums`, return the smallest missing positive integer.
+Given an unsorted integer array `nums`. Return the _smallest positive integer_ that is _not present_ in `nums`.
 
-You must implement an algorithm that runs in `O(n)` time and uses constant extra space.
+You must implement an algorithm that runs in `O(n)` time and uses `O(1)` auxiliary space.
 
 **Example 1:**
 
 **Input:** nums = [1,2,0]
 
 **Output:** 3
 
-**Explanation:** The numbers in the range [1,2] are all in the array.
+**Explanation:** The numbers in the range [1,2] are all in the array. 
 
 **Example 2:**
 
 **Input:** nums = [3,4,-1,1]
 
 **Output:** 2
 
-**Explanation:** 1 is in the array but 2 is missing.
+**Explanation:** 1 is in the array but 2 is missing. 
 
 **Example 3:**
 
 **Input:** nums = [7,8,9,11,12]
 
 **Output:** 1
 
-**Explanation:** The smallest positive integer 1 is missing.
+**Explanation:** The smallest positive integer 1 is missing. 
 
 **Constraints:**
 

src/main/go/g0001_0100/s0041_first_missing_positive/solution.go

@@ -1,42 +1,37 @@
 package s0041_first_missing_positive
 
 // #Hard #Top_100_Liked_Questions #Top_Interview_Questions #Array #Hash_Table #Udemy_Arrays
-// #Big_O_Time_O(n)_Space_O(n) #2024_03_12_Time_39_ms_(92.34%)_Space_7.8_MB_(79.50%)
+// #Big_O_Time_O(n)_Space_O(n) #2025_05_03_Time_0_ms_(100.00%)_Space_9.49_MB_(95.52%)
 
 func firstMissingPositive(nums []int) int {
 	for i := 0; i < len(nums); i++ {
 		if nums[i] < 0 {
 			nums[i] = 0
 		}
 	}
-
 	for i := 0; i < len(nums); i++ {
 		val := abs(nums[i])
 		if val > 0 && val <= len(nums) {
 			if nums[val-1] > 0 {
 				nums[val-1] = nums[val-1] * -1
 				continue
 			}
-
 			if nums[val-1] == 0 {
 				nums[val-1] = -1 * (len(nums) + 1)
 			}
 		}
 	}
-
 	for i := 1; i <= len(nums); i++ {
 		if nums[i-1] >= 0 {
 			return i
 		}
 	}
-
 	return len(nums) + 1
 }
 
 func abs(n int) int {
 	if n >= 0 {
 		return n
 	}
-
 	return n * -1
 }

src/main/go/g0001_0100/s0042_trapping_rain_water/solution.go

@@ -3,7 +3,7 @@ package s0042_trapping_rain_water
 // #Hard #Top_100_Liked_Questions #Top_Interview_Questions #Array #Dynamic_Programming #Two_Pointers
 // #Stack #Monotonic_Stack #Dynamic_Programming_I_Day_9 #Udemy_Two_Pointers
 // #Top_Interview_150_Array/String #Big_O_Time_O(n)_Space_O(1)
-// #2024_03_12_Time_3_ms_(99.42%)_Space_5.4_MB_(97.01%)
+// #2025_05_03_Time_0_ms_(100.00%)_Space_7.85_MB_(82.53%)
 
 func trap(height []int) int {
 	if height == nil {

src/main/go/g0001_0100/s0045_jump_game_ii/readme.md

@@ -2,29 +2,31 @@
 
 Medium
 
-Given an array of non-negative integers `nums`, you are initially positioned at the first index of the array.
+You are given a **0-indexed** array of integers `nums` of length `n`. You are initially positioned at `nums[0]`.
 
-Each element in the array represents your maximum jump length at that position.
+Each element `nums[i]` represents the maximum length of a forward jump from index `i`. In other words, if you are at `nums[i]`, you can jump to any `nums[i + j]` where:
 
-Your goal is to reach the last index in the minimum number of jumps.
+*   `0 <= j <= nums[i]` and
+*   `i + j < n`
 
-You can assume that you can always reach the last index.
+Return _the minimum number of jumps to reach_ `nums[n - 1]`. The test cases are generated such that you can reach `nums[n - 1]`.
 
 **Example 1:**
 
 **Input:** nums = [2,3,1,1,4]
 
 **Output:** 2
 
-**Explanation:** The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
+**Explanation:** The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index. 
 
 **Example 2:**
 
 **Input:** nums = [2,3,0,1,4]
 
-**Output:** 2
+**Output:** 2 
 
 **Constraints:**
 
 *   <code>1 <= nums.length <= 10<sup>4</sup></code>
-*   `0 <= nums[i] <= 1000`
+*   `0 <= nums[i] <= 1000`
+*   It's guaranteed that you can reach `nums[n - 1]`.

src/main/go/g0001_0100/s0045_jump_game_ii/solution.go

@@ -3,7 +3,7 @@ package s0045_jump_game_ii
 // #Medium #Top_100_Liked_Questions #Array #Dynamic_Programming #Greedy
 // #Algorithm_II_Day_13_Dynamic_Programming #Dynamic_Programming_I_Day_4
 // #Top_Interview_150_Array/String #Big_O_Time_O(n)_Space_O(1)
-// #2024_03_12_Time_6_ms_(96.98%)_Space_6.3_MB_(34.53%)
+// #2025_05_03_Time_0_ms_(100.00%)_Space_7.78_MB_(85.74%)
 
 func jump(nums []int) int {
 	length := 0

src/main/go/g0001_0100/s0046_permutations/readme.md

@@ -2,25 +2,25 @@
 
 Medium
 
-Given an array `nums` of distinct integers, return _all the possible permutations_. You can return the answer in **any order**.
+Given an array `nums` of distinct integers, return all the possible permutations. You can return the answer in **any order**.
 
 **Example 1:**
 
 **Input:** nums = [1,2,3]
 
-**Output:** [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
+**Output:** [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] 
 
 **Example 2:**
 
 **Input:** nums = [0,1]
 
-**Output:** [[0,1],[1,0]]
+**Output:** [[0,1],[1,0]] 
 
 **Example 3:**
 
 **Input:** nums = [1]
 
-**Output:** [[1]]
+**Output:** [[1]] 
 
 **Constraints:**
 

src/main/go/g0001_0100/s0046_permutations/solution.go

@@ -3,7 +3,7 @@ package s0046_permutations
 // #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Backtracking
 // #Algorithm_I_Day_11_Recursion_Backtracking #Level_2_Day_20_Brute_Force/Backtracking
 // #Udemy_Backtracking/Recursion #Top_Interview_150_Backtracking #Big_O_Time_O(n*n!)_Space_O(n+n!)
-// #2024_03_12_Time_0_ms_(100.00%)_Space_2.7_MB_(51.79%)
+// #2025_05_03_Time_0_ms_(100.00%)_Space_4.59_MB_(85.96%)
 
 func permute(nums []int) [][]int {
 	if len(nums) == 0 {

src/main/go/g0001_0100/s0048_rotate_image/solution.go

@@ -3,7 +3,7 @@ package s0048_rotate_image
 // #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Math #Matrix
 // #Data_Structure_II_Day_3_Array #Programming_Skills_II_Day_7 #Udemy_2D_Arrays/Matrix
 // #Top_Interview_150_Matrix #Big_O_Time_O(n^2)_Space_O(1)
-// #2024_03_12_Time_0_ms_(100.00%)_Space_2.3_MB_(75.46%)
+// #2025_05_03_Time_0_ms_(100.00%)_Space_4.06_MB_(97.99%)
 
 func rotate(matrix [][]int) {
 	n := len(matrix)

src/main/go/g0401_0500/s0494_target_sum/solution.go

@@ -1,7 +1,7 @@
 package s0494_target_sum
 
 // #Medium #Array #Dynamic_Programming #Backtracking #Big_O_Time_O(n*(sum+s))_Space_O(n*(sum+s))
-// #2024_03_12_Time_7_ms_(80.54%)_Space_6.6_MB_(40.47%)
+// #2025_04_30_Time_4_ms_(82.40%)_Space_8.36_MB_(41.63%)
 
 import "math"
 
@@ -11,25 +11,21 @@ func findTargetSumWays(nums []int, s int) int {
 	for _, num := range nums {
 		sum += num
 	}
-
 	if s > sum || (sum+s)%2 != 0 {
 		return 0
 	}
-
 	dp := make([][]int, (sum+s)/2+1)
 	for i := range dp {
 		dp[i] = make([]int, len(nums)+1)
 	}
 	dp[0][0] = 1
-
 	for i := 0; i < len(nums); i++ {
 		if nums[i] == 0 {
 			dp[0][i+1] = dp[0][i] * 2
 		} else {
 			dp[0][i+1] = dp[0][i]
 		}
 	}
-
 	for i := 1; i < len(dp); i++ {
 		for j := 0; j < len(nums); j++ {
 			dp[i][j+1] += dp[i][j]
@@ -38,6 +34,5 @@ func findTargetSumWays(nums []int, s int) int {
 			}
 		}
 	}
-
 	return dp[(sum+s)/2][len(nums)]
 }