added some solution between 3000-3010

dsa-solutions/lc-solutions/3000-3099/3000-Maximum area of diagonal rectangle.md

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+---
+id: 3000
+title: Maximum area of longest diagonal rectangle (Leetcode)
+sidebar_label: 3000-Maximum area of longest diagonal rectangle
+description: Solution of Maximum area of longest diagonal rectangle
+---
+
+## Problem Description
+
+| Problem Statement                                                                                                                   | Solution Link | LeetCode Profile |
+| :---------------------------------------------------------------------------------------------------------------------------------- | :------------ | :--------------- |
+| [Maximum area of longest diagonal rectangle](https://leetcode.com/problems/maximum-area-of-longest-diagonal-rectangle/description/) |
+
+## Problem Description
+
+You are given a 2D 0-indexed integer array dimensions.
+
+For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.
+
+Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.
+
+### Examples
+
+#### Example 1
+
+- **Input:** $dimensions = [[9,3],[8,6]]$
+- **Output:** $48$
+- **Explanation:** $For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
+For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
+So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48..$
+
+#### Example 2
+
+- **Input:** $dimensions = [[3,4],[4,3]]$
+- **Output:** $12$
+- **Explanation:**$ Length of diagonal is the same for both which is 5, so maximum area = 12$.
+
+### Constraints
+
+- $1 <= dimensions.length <= 100$
+- $dimensions[i].length == 2$
+- $1 <= dimensions[i][0], dimensions[i][1] <= 100$
+
+### Intuition
+
+The problem requires finding the rectangle with the maximum diagonal length and, in case of a tie, returning the one with the maximum area.
+To achieve this, we need to calculate the diagonal length for each rectangle and compare it with the current maximum diagonal length.
+If the diagonal length is greater or equal (in case of a tie), we update the maximum diagonal length and check if the area of the current rectangle is greater than the maximum area. If so, we update the maximum area as well.
+
+### Approach
+
+The code uses a loop to iterate through each rectangle in the given dimensions vector.
+It calculates the square of the diagonal length for each rectangle and compares it with the previous maximum diagonal length (diag).
+If the current rectangle has a longer or equal diagonal length, it updates the maximum area (area) accordingly.
+The area variable keeps track of the maximum area among rectangles with the longest diagonal.
+Finally, the function returns the maximum area among rectangles with the longest diagonal.
+
+### Solution Code
+
+#### Python
+
+```py
+  class Solution:
+    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:
+        a,b = max(dimensions, key = lambda x: (x[0]*x[0]+x[1]*x[1], (x[0]*x[1])))
+      return a*b
+```
+
+#### Java
+
+```java
+
+  public int areaOfMaxDiagonal(int[][] dimensions) {
+        int diagSq = 0, area = 0;
+        for (int[] d : dimensions) {
+            int ds = d[0] * d[0] + d[1] * d[1];
+            int ar = d[0] * d[1];
+            if (ds > diagSq) {
+                area = ar;
+                diagSq = ds;
+            }
+            else if (ds == diagSq && ar > area) area = ar;
+        }
+        return area;
+    }
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int areaOfMaxDiagonal(vector<vector<int>>& dimensions) {
+
+        double max_diagonal = 0;
+        int max_area = 0;
+
+        for (const auto& rectangle : dimensions) {
+            int length = rectangle[0];
+            int width = rectangle[1];
+
+            double diagonal_length = sqrt(pow(length, 2) + pow(width, 2));
+
+            if (diagonal_length > max_diagonal || (diagonal_length == max_diagonal && length * width > max_area)) {
+                max_diagonal = diagonal_length;
+                max_area = length * width;
+            }
+        }
+
+        return max_area;
+
+    }
+};
+```
+
+### Conclusion
+
+At last we can say that we can calculate maximum area of diagonal rectangle using timecomplexity of o(n) and spacecomplexity of o(1) using a simple for loop and some variable

dsa-solutions/lc-solutions/3000-3099/3001-Minimum moves to capture queen

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+---
+id: Minimum Moves to Capture The Queen
+title: Minimum Moves to Capture The Queen
+sidebar_label: 3001 - Minimum Moves to Capture The Queen
+tags:
+  - Array
+description: "This is a solution to the Minimum Moves to Capture The Queen problem on LeetCode."
+---
+
+## Problem Description
+
+There is a 1-indexed 8 x 8 chessboard containing 3 pieces.
+
+You are given 6 integers a, b, c, d, e, and f where:
+
+(a, b) denotes the position of the white rook.
+(c, d) denotes the position of the white bishop.
+(e, f) denotes the position of the black queen.
+Given that you can only move the white pieces, return the minimum number of moves required to capture the black queen.
+
+Note that:
+
+Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces.
+Bishops can move any number of squares diagonally, but cannot jump over other pieces.
+A rook or a bishop can capture the queen if it is located in a square that they can move to.
+The queen does not move.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
+Output: 2
+Explanation: We can capture the black queen in two moves by moving the white rook to (1, 3) then to (2, 3).
+It is impossible to capture the black queen in less than two moves since it is not being attacked by any of the pieces at the beginning.
+
+```
+
+**Example 2:**
+
+```
+Input: a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
+Output: 1
+Explanation: We can capture the black queen in a single move by doing one of the following:
+- Move the white rook to (5, 2).
+- Move the white bishop to (5, 2).
+
+```
+
+### Constraints
+
+- `1 <= a, b, c, d, e, f <= 8`
+- `No two pieces are on the same square.`
+
+## Solution for Minimum Moves to Capture The Queen Problem
+
+### Intuition
+
+Same Row or Column:
+If the queen is in the same row (a == e) or the same column (b == f), it means it can be captured in one move.
+
+Same Diagonal:
+If the absolute difference between the columns (abs(c - e)) is equal to the absolute difference between the rows (abs(d - f)), it means the queen is on the same diagonal and can be captured in one move.
+
+Otherwise:
+If none of the above conditions are met, it implies that the queen cannot be captured in one move.
+
+Pieces in Between
+For suppose a rook is standing in the way of bishop and vice-versa the min no of moves is two , to check whether the piece in between i have used the manhattan distance.
+
+### Approach
+
+Check for Same Row, Column, or Diagonal:
+If the queen is in the same row, column, or diagonal as the starting point, calculate the Manhattan distance.
+
+Check for Obstacles:
+While calculating the Manhattan distance, inspect the squares between the starting point and the queen's position.
+If any square along the path contains another piece, consider it an obstacle.
+
+Minimum Moves Calculation:
+Assess the minimum number of moves based on the presence of obstacles.
+If no obstacles are present, it's likely a direct capture in one move.
+If obstacles are encountered, it might require at least two moves to maneuver around them.
+
+Handle Piece-Specific Scenarios:
+Adjust the approach based on the specific pieces involved.
+For a rook, obstacles in the same row or column are significant.
+For a bishop, obstacles along diagonals need consideration.
+
+Return Minimum Moves:
+Return the calculated minimum number of moves.
+
+### Solution Code
+
+#### Python
+
+```py
+class Solution:
+    def minMovesToCaptureTheQueen(self, a: int, b: int, c: int, d: int, e: int, f: int) -> int:
+        # rock and queen in the same row or col
+        if a == e: # same row
+            if a == c and (d - b) * (d - f) < 0: # bishop on the same row and between rock and queen
+                return 2
+            else:
+                return 1
+        elif b == f: # same col
+            if b == d and (c - a) * (c - e) < 0:
+                return 2
+            else:
+                return 1
+        # bishop and queen in the same diagonal
+        elif c - e == d - f: # \ diagonal
+            if a - e == b - f and (a - c) * (a - e) < 0:
+                return 2
+            else:
+                return 1
+        elif c - e == f - d: # / diagonal
+            if a - e == f - b and (a - c) * (a - e) < 0:
+                return 2
+            else:
+                return 1
+        return 2
+
+
+
+```
+
+#### Java
+
+```java
+=public int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
+	// ----- Rook -----
+	if(a==e){
+		if(a==c && ((b<d && d<f) || (f<d && d<b))) {
+			return 2;
+		}else return 1;
+
+	}else if(b==f) {
+		if(b==d && ((a<c && c<e) || (e<c && c<a))){
+			return 2;
+		}else return 1;
+	}
+
+
+	// ----- Bishop -----
+	// flag -> true means rook is between bishop and queen - so we need to move rook
+	boolean flag = false;
+	for(int i=c, j=d; i>0&&j>0; i--, j--){
+		if(i==a && j==b) flag = true;
+		else if(i==e && j==f) {
+			if(flag) return 2;
+			else return 1;
+		}
+	}
+
+	flag =false;
+	for(int i=c, j=d; i>0&&j<9; i--, j++){
+		if(i==a && j==b) flag = true;
+		else if(i==e && j==f) {
+			if(flag) return 2;
+			else return 1;
+		}
+	}
+
+	flag = false;
+	for(int i=c, j=d; i<9&&j<9; i++, j++){
+		if(i==a && j==b) flag = true;
+		else if(i==e && j==f) {
+			if(flag) return 2;
+			else return 1;
+		}
+	}
+
+	flag = false;
+	for(int i=c, j=d; i<9&&j>0; i++, j--){
+		if(i==a && j==b) flag = true;
+		else if(i==e && j==f) {
+			if(flag) return 2;
+			else return 1;
+		}
+	}
+
+	// for every condition Rook can capture the queen in 2 steps
+	return 2;
+}
+
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
+        if(a == e){
+            if(c == e){
+                int d0 = abs(b-f) ;
+                int d1 = abs(b-d) ;
+                int d2 = abs(d-f);
+                if(d0 == d1+d2)
+                    return 2;
+                else
+                    return 1;
+            }
+            else
+                return 1;
+        }
+        else if(b == f){
+            if(d == f){
+                int d0 = abs(a-e) ;
+                int d1 = abs(a-c);
+                int d2 = abs(c-e) ;
+                if(d0 == d1+d2)
+                    return 2;
+                else
+                    return 1;
+            }
+            else
+                return 1;
+
+        }
+        else if(abs(c-e) == abs(d - f)){
+            if(abs(a-c) == abs(b-d)) {
+                int d0 = abs(c-e) + abs(d-f);
+                int d1 = abs(a-c) + abs(b-d);
+                int d2 = abs(e-a) + abs(f-b);
+                if(d0 == d1+d2)
+                    return 2;
+                else
+                    return 1;
+            }
+            else
+                return 1;
+        }
+        else
+            return 2;
+
+    }
+};
+```
+
+### Conclusion
+
+THis code efficectively calculates the minimum number of moves using constant space and constant time
+
+- **LeetCode Problem**: [Minimum Moves to Capture The Queen](https://leetcode.com/problems/minimum-moves-to-capture-the-queen/description/)
+
+- **Solution Link**: [LeetCode Solution](https://leetcode.com/problems/minimum-moves-to-capture-the-queen/solutions/4577719/beats-100-time-complexity-o-1)