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2Q) You are given a permutation P of length N. This permutation represents a graph of N nodes where for each node i from 1 to N there is an ongoing edge from that node to node P[i]. A permutation is an array of length N, consisting of each of the integers from 1 to N in some order. The longest path in the graph is the path that satisfies the following conditions: تھا •⁠ ⁠It starts at some node U and ends at some node V. It visits each node no more than once. •⁠ ⁠Among all the possible paths, it's the longest one. Find the total number of possible pairs of indices of the permutation (i, j), such that: •⁠ ⁠If P[i] and P[j] are swapped, then the resulting graph has the maximum possible longest path among all the possible swaps. Since the answer is very large, print it modulo 109+7. Input Format The first line contains an integer, N, denoting the number of elements in P. Each line i of the N subsequent lines (where 0 ≤ i < N) contains an integer describing P[i]. Sample Test Cases Case 1 Input: 3 1 2 3 Output: 3 Explanation: Given N 3, P= [1, 2, 3]. Here, if we swap "P[1]" and "P[2]" we will get P = [2, 1, 3]", node "1" can go to node "2" and we can consider that the longest path, and also node "2" can go to node "1", we also can generate the following two permutations: Case 2 Input: 6 2 3 1 S 6 4 Output: 9 Explanation: Given N 6, P [2, 3, 1, 5, 6, 4]. Here, we have two cycles, the nodes in the first cycle are "[1, 2, 3]" and the nodes in the second cycle are "[4, 5, 6]". We can show that if we swap any of the first three elements in the permutation with any element from the last three elements, the cycles will be merged, so the answer is "339". Case 3 Input: 7 2 3 1 5 4 7 6 Output: 12 Explanation: Given N 7, P [2, 3, 1, 5, 4, 7, 6]. Here, if we swap any element from the first three elements with any element from the last four elements we'll get a longest path of length "6", and from that the number of swaps to obtain that length (the maximum possible reachable length of the graph) is "3 * 4 = 12".

1Ans) java code import java.util.ArrayList; import java.util.List; public class Solution { public static int getAns(int N, List> A) { // This will store the maximum beauty sum of two non-intersecting squares int maxSumOfBeauties = 0; // Precompute the beauty of all sub-grids for quick lookup int[][] beauty = new int[N][N]; for (int row = 0; row < N; row++) { for (int col = 0; col < N; col++) { // Calculate beauty for sub-grid starting at (row, col) int minVal = Integer.MAX_VALUE; int maxVal = Integer.MIN_VALUE; for (int i = row; i < N; i++) { for (int j = col; j < N; j++) { minVal = Math.min(minVal, A.get(i).get(j)); maxVal = Math.max(maxVal, A.get(i).get(j)); beauty[i][j] = maxVal - minVal; } } } } // Try all possible pairs of non-intersecting sub-grids for (int r1 = 0; r1 < N; r1++) { for (int c1 = 0; c1 < N; c1++) { for (int r2 = 0; r2 < N; r2++) { for (int c2 = 0; c2 < N; c2++) { if (r1 != r2 && c1 != c2) { // Ensure the sub-grids do not share any rows or columns int beauty1 = beauty[r1][c1]; int beauty2 = beauty[r2][c2]; maxSumOfBeauties = Math.max(maxSumOfBeauties, beauty1 + beauty2); } } } } } return maxSumOfBeauties; } public static void main(String[] args) { // Example usage: List> grid = new ArrayList<>(); grid.add(List.of(2, 3)); grid.add(List.of(2, 3)); int N = 2; System.out.println(getAns(N, grid)); // Output should be 0 } }

1Ans) java code import java.util.ArrayList; import java.util.List; public class Solution { public static int getAns(int N, List> A) { // This will store the maximum beauty sum of two non-intersecting squares int maxSumOfBeauties = 0; // Precompute the beauty of all sub-grids for quick lookup int[][] beauty = new int[N][N]; for (int row = 0; row < N; row++) { for (int col = 0; col < N; col++) { // Calculate beauty for sub-grid starting at (row, col) int minVal = Integer.MAX_VALUE; int maxVal = Integer.MIN_VALUE; for (int i = row; i < N; i++) { for (int j = col; j < N; j++) { minVal = Math.min(minVal, A.get(i).get(j)); maxVal = Math.max(maxVal, A.get(i).get(j)); beauty[i][j] = maxVal - minVal; } } } } // Try all possible pairs of non-intersecting sub-grids for (int r1 = 0; r1 < N; r1++) { for (int c1 = 0; c1 < N; c1++) { for (int r2 = 0; r2 < N; r2++) { for (int c2 = 0; c2 < N; c2++) { if (r1 != r2 && c1 != c2) { // Ensure the sub-grids do not share any rows or columns int beauty1 = beauty[r1][c1]; int beauty2 = beauty[r2][c2]; maxSumOfBeauties = Math.max(maxSumOfBeauties, beauty1 + beauty2); } } } } } return maxSumOfBeauties; } public static void main(String[] args) { // Example usage: List> grid = new ArrayList<>(); grid.add(List.of(2, 3)); grid.add(List.of(2, 3)); int N = 2; System.out.println(getAns(N, grid)); // Output should be 0 } }

Q1: Two Square minimax You are given a square grid A of size N x N. Your want to choose two non-intersecting square sub-grids from the grid such that they have no common row or column The objective is to maximize the sum of beauties of both square sub-grids. The beauty of a sub-grid is defined as the difference betwee the maximum value and the minimum value within that sub- grid. tikqupta2s Find the maximum possible sum of beauties for the sel two sub-grids. Input Format The next line contains an integer, N, denoting the number o columns in A. Each line i of the N subsequent lines (where 0 ≤ i < N) com space separated integers each describing the row A[i]. Constraints 2 <= N <= 250 1 <= A[i][j] <= 10^5 Sample Test Cases Case 1 Input: 2 23 23 Output: 0 Explanation: Given N 2, A = [[2, 3], [2, 3]]. We take cell (1, 2) and cell (2, 1) and the answer will be 0 Public class solution { Public static int get_ans(int N, List> A}}

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