Merge pull request #3939 from ImmidiSivani/leetcode-648

added solution to leetcode 648

Author: ajay-dhangar

dsa-solutions/gfg-solutions/Easy problems/Square-Root.md

@@ -3,8 +3,8 @@ id: square-root
 title: Square Root
 sidebar_label: Square-Root
 tags:
-- Math
-- Binary Search
+  - Math
+  - Binary Search
 description: "This document provides solutions to the problem of finding the Square Root of an integer."
 ---
 
@@ -38,6 +38,7 @@ You don't need to read input or print anything. The task is to complete the func
 **Expected Auxiliary Space:** $O(1)$
 
 **Constraints**
+
 - `1 ≤ x ≤ 10^7`
 
 ## Solution
@@ -128,25 +129,27 @@ public:
 
 ```javascript
 class Solution {
-    floorSqrt(x) {
-        if (x === 0 || x === 1) {
-            return x;
-        }
-        let start = 1, end = x, ans = 0;
-        while (start <= end) {
-            let mid = Math.floor((start + end) / 2);
-            if (mid * mid === x) {
-                return mid;
-            }
-            if (mid * mid < x) {
-                start = mid + 1;
-                ans = mid;
-            } else {
-                end = mid - 1;
-            }
-        }
-        return ans;
+  floorSqrt(x) {
+    if (x === 0 || x === 1) {
+      return x;
+    }
+    let start = 1,
+      end = x,
+      ans = 0;
+    while (start <= end) {
+      let mid = Math.floor((start + end) / 2);
+      if (mid * mid === x) {
+        return mid;
+      }
+      if (mid * mid < x) {
+        start = mid + 1;
+        ans = mid;
+      } else {
+        end = mid - 1;
+      }
     }
+    return ans;
+  }
 }
 ```
 
@@ -155,25 +158,27 @@ class Solution {
 
 ```typescript
 class Solution {
-    floorSqrt(x: number): number {
-        if (x === 0 || x === 1) {
-            return x;
-        }
-        let start = 1, end = x, ans = 0;
-        while (start <= end) {
-            let mid = Math.floor((start + end) / 2);
-            if (mid * mid === x) {
-                return mid;
-            }
-            if (mid * mid < x) {
-                start = mid + 1;
-                ans = mid;
-            } else {
-                end = mid - 1;
-            }
-        }
-        return ans;
+  floorSqrt(x: number): number {
+    if (x === 0 || x === 1) {
+      return x;
+    }
+    let start = 1,
+      end = x,
+      ans = 0;
+    while (start <= end) {
+      let mid = Math.floor((start + end) / 2);
+      if (mid * mid === x) {
+        return mid;
+      }
+      if (mid * mid < x) {
+        start = mid + 1;
+        ans = mid;
+      } else {
+        end = mid - 1;
+      }
     }
+    return ans;
+  }
 }
 ```
 
@@ -185,4 +190,4 @@ class Solution {
 The provided solutions efficiently find the floor value of the square root of a given integer `x` using binary search. This approach ensures a time complexity of $ O(log N) and an auxiliary space complexity of $O(1)$. The algorithms are designed to handle large values of `x` up to 10^7 efficiently without relying on built-in square root functions.
 
 **Time Complexity:** $O(log N)$
-**Auxiliary Space:** $O(1)$
+**Auxiliary Space:** $O(1)$

dsa-solutions/lc-solutions/0200-0299/0240-search-a-2d-matrix-II.md

@@ -69,14 +69,15 @@ Output: false
 ### Intuition and Approach
 
 To search for a value in this matrix efficiently, we can utilize the properties of the matrix. Since the matrix is sorted both row-wise and column-wise, we can start our search from the top-right corner of the matrix. From here, we have two options:
+
 1. If the target is greater than the current value, move downwards.
 2. If the target is less than the current value, move leftwards.
 
 This approach ensures that we eliminate one row or one column in each step, leading to an efficient search.
 
 <Tabs>
  <tabItem value="Greedy Search" label="Greedy Search">
-  
+
 ### Approach: Greedy Search
 
 By leveraging the sorted properties of the matrix, we can search for the target value efficiently using a greedy approach. This involves starting from the top-right corner and adjusting our search direction based on the current value.
@@ -108,21 +109,16 @@ const matrix = [
   [2, 5, 8, 12, 19],
   [3, 6, 9, 16, 22],
   [10, 13, 14, 17, 24],
-  [18, 21, 23, 26, 30]
+  [18, 21, 23, 26, 30],
 ];
 const target = 5;
 const result = searchMatrix(matrix, target);
 
 return (
   <div>
     <p>
-      <b>Input:</b> matrix = [
-        [1, 4, 7, 11, 15],
-        [2, 5, 8, 12, 19],
-        [3, 6, 9, 16, 22],
-        [10, 13, 14, 17, 24],
-        [18, 21, 23, 26, 30]
-      ], target = 5
+      <b>Input:</b> matrix = [ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9,
+      16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ], target = 5
     </p>
     <p>
       <b>Output:</b> {result ? "true" : "false"}
@@ -182,7 +178,7 @@ return (
     ```
 
   </TabItem>
-  <TabItem value="Python" label="Python"> 
+  <TabItem value="Python" label="Python">
   <SolutionAuthor name="@aryansh-patel"/>
    ```python
     def searchMatrix(matrix: List[List[int]], target: int) -> bool:
@@ -259,7 +255,7 @@ return (
     };
     ```
 
-  </TabItem>  
+  </TabItem>
 </Tabs>
 
 #### Complexity Analysis
@@ -269,7 +265,8 @@ return (
 
 - The time complexity is linear in terms of the dimensions of the matrix. Each step eliminates either a row or a
 
- column, leading to a linear time complexity of $O(m + n)$.
+column, leading to a linear time complexity of $O(m + n)$.
+
 - The space complexity is constant because we only use a few extra variables regardless of the matrix size.
 
 </tabItem>
@@ -287,16 +284,16 @@ This solution leverages the matrix's properties to reduce the search space effic
 
   <TabItem value="en" label="English">
 
-  ---
-  
+---
+
     <Tabs>
       <TabItem value="javascript" label="JavaScript">
         <LiteYouTubeEmbed
           id="mH0X4jxrQjY"
           params="autoplay=1&autohide=1&showinfo=0&rel=0"
           title="Search a 2D Matrix II Problem Explanation | Search a 2D Matrix II Solution"
           poster="maxresdefault"
-          webp 
+          webp
         />
       </TabItem>
 
@@ -306,7 +303,7 @@ This solution leverages the matrix's properties to reduce the search space effic
           params="autoplay=1&autohide=1&showinfo=0&rel=0"
           title="Search a 2D Matrix II Problem Explanation | Search a 2D Matrix II Solution"
           poster="maxresdefault"
-          webp 
+          webp
         />
       </TabItem>
       <TabItem value="java" label="Java">
@@ -315,7 +312,7 @@ This solution leverages the matrix's properties to reduce the search space effic
           params="autoplay=1&autohide=1&showinfo=0&rel=0"
           title="Search a 2D Matrix II Problem Explanation | Search a 2D Matrix II Solution"
           poster="maxresdefault"
-          webp 
+          webp
         />
       </TabItem>
     </Tabs>
@@ -328,7 +325,7 @@ This solution leverages the matrix's properties to reduce the search space effic
           params="autoplay=1&autohide=1&showinfo=0&rel=0"
           title="Search a 2D Matrix II Problem Explanation | Search a 2D Matrix II Solution"
           poster="maxresdefault"
-          webp 
+          webp
         />
   </TabItem>