Kodeday

  You are given an array (zero-indexed) of   N   non-negative integers,   A 0 ,   A 1   ,…,   A N-1 . Find the minimum sub-array  A l ,  A l+1 ,…,  A r  so if we sort(in ascending order) that sub-array, then the whole array should get sorted. If  A  is already sorted, output  -1 . Example : Input 1: A = [1, 3, 2, 4, 5] Return: [1, 2] Input 2: A = [1, 2, 3, 4, 5] Return: [-1] In the above example(Input 1), if we sort the subarray  A 1 ,  A 2 , then whole array  A  should get sorted. ut Solution: class Solution:     # @param A : list of integers     # @return a list of integers     def subUnsort(self, A):         s_a=sorted(A)         if A==s_a:             return [-1]         l=len(A)         minI=-1         maxI=l       ...

  Problem Description Given an array  A  of integers, find the maximum of  j - i  subjected to the constraint of  A[i] <= A[j] . Input Format First and only argument is an integer array A. Output Format Return an integer denoting the maximum value of j - i; Example Input Input 1: A = [3, 5, 4, 2] Example Output Output 1: 2 Example Explanation Explanation 1: Maximum value occurs for pair (3, 4). Solution: class Solution:     # @param A : tuple of integers     # @return an integer     def maximumGap(self, A):         n=len(A)             R = [0] * n              R[n - 1] = A[n - 1]          for j in range(n - 2, -1, -1):              R[j] = max(A[j], R[j + 1]);               i = 0;j=0         diff = -1     ...

  You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). You need to do this in place. Note that if you end up using an additional array, you will only receive partial score. Example: If the array is [ [1, 2], [3, 4] ] Then the rotated array becomes: [ [3, 1], [4, 2] ] Solution: class Solution:     # @param A : list of list of integers     # @return the same list modified     def rotate(self, A):         l = len(A)         for i in range(l//2):             A[i], A[l-i-1] = A[l-i-1], A[i]         ans = [[A[j][i] for j in range(l)] for i in range(len(A[0]))]         return ans Screenshot:

  Given a read-only array of n + 1 integers between 1 and n, find one number that repeats in linear time using less than O(n) space and traversing the stream sequentially O(1) times. Sample Input: [3 4 1 4 1] Sample Output: 1 If there are multiple possible answers ( like in the sample case above ), output any one. If there is no duplicate, output -1 Solution class Solution:     # @param A : tuple of integers     # @return an integer     def repeatedNumber(self, A):         l_A=len(A)         s_A=set(A)         d={}         l_s=len(s_A)         if l_A==l_s:             return -1         for i in A:             if i in d:                 d[i]+=1             else:             ...