Microsoft Corporation SDE Interview Questions and Answers

Find an index in array where sum of elements on left side is equal to sum of elements on right side.

• Loop through array and calculate sum of all elements

• Then loop through array again and check if sum of elements on left side is equal to sum of elements on right side

• Return the index if found, else return -1

Find minimum number of jumps required to reach end of array with max moves in right direction at each index.

• Use dynamic programming approach to solve the problem

• Maintain an array to store minimum number of jumps required to reach each index

• Traverse the array and update the minimum number of jumps for each index

• Return the minimum number of jumps required to reach the end of array

Tournament tree is a binary tree where each node represents a match between two players/teams.

• Tournament tree is used in sports tournaments to determine the winner.

• The leaf nodes represent the players/teams and the internal nodes represent the matches.

• The winner of each match is the node with the higher score.

• Code can be written using a recursive function to traverse the tree and determine the winner.

Find 2 maximum elements in array in O(n+log(n)) comparison

• Use divide and conquer approach

• Divide array into two halves

• Recursively find max elements in each half

• Compare the two max elements to find the overall max elements

Find row with maximum 1s in a sorted 2D array of 0s and 1s.

• Use binary search to find the first occurrence of 1 in each row.

• Calculate the number of 1s in each row using the index found in step 1.

• Keep track of the row with maximum number of 1s.

• Time complexity: O(m log n), where m is the number of rows and n is the number of columns.

Efficiently check if a number is present in an array where each element differs by 1.

• Use binary search to find the element in the array

• Calculate the difference between the middle element and the first element

• If the difference is equal to the index of the middle element minus the index of the first element, then the left half of the array is consecutive

• If the difference is not equal, then the right half of the array is ...read more

Microsoft Excel

• Add more chart types (p1)

• Improve collaboration features (p2)

• Enhance data analysis tools (p3)

• Integrate with more external data sources (p4)

• Improve error handling and debugging (p5)

Program to rearrange shuffled 16 squares to get original big square

• Create a 4x4 matrix to represent the big square and fill it with shuffled squares

• Loop through the matrix and check if each square is in the correct position

• If a square is not in the correct position, swap it with the square in the correct position

• Repeat until all squares are in the correct position

Find pairs from sorted arrays a and b that sum to elements in sorted array c.

• Use a two-pointer technique to traverse arrays a and b.

• For each element in c, check if a[i] + b[j] equals c[k].

• Example: a = [1, 2], b = [3, 4], c = [4, 5] results in pairs (1, 3).

• Start with pointers at the beginning of a and b, and adjust based on the sum.

The first declaration casts i to int pointer and dereferences it, while the second declaration casts the address of i to int pointer and dereferences it.

• The first declaration assumes i is already an int pointer.

• The second declaration takes the address of i and casts it to an int pointer.

• The first declaration may cause a segmentation fault if i is not an int pointer.

• The second declaration may cause unexpected behavior i...read more

Minimum comparisons needed to find median in unsorted array of size n

• For odd n, median is the middle element. For even n, median is the average of middle two elements

• Minimum comparisons needed for odd n is (n-1)/2. For even n, it is n/2

• Various sorting algorithms can be used to find median in an unsorted array

Time complexity to find an element in an alternatively sorted array.

• The time complexity will be O(log n) using binary search.

• Check the middle element and compare with the target element.

• If the middle element is greater than the target, search in the left half of the array.

• If the middle element is smaller than the target, search in the right half of the array.

• Repeat until the target element is found or the array is exha...read more