Sort Array Based on Parity Values in Python

Arnab Chakraborty
Updated on 21-Oct-2020 11:07:37

372 Views

Suppose, we have an array A with few integers. We have to sort the numbers as even then odd. So put the even numbers at first, then the odd numbers. So if the array is like A = [1, 5, 6, 8, 7, 2, 3], then the result will be like [6, 8, 2, 1, 5, 7, 3]To solve this, we will follow these steps −set i := 0 and j := 0while j < size of arrif arr[j] is even, thenswap arr[i] and arr[j], increase i by 1increase j by 1return arrLet us see the following implementation to get ... Read More

Find Smallest Intersecting Element of Each Row in a Matrix in Python

Arnab Chakraborty
Updated on 21-Oct-2020 11:06:05

178 Views

Suppose we have a 2D matrix where each row is sorted in ascending order. We have to find the smallest number that exists in every row. If there's no such result, then return −1.So, if the input is like23551010135then the output will be 5To solve this, we will follow these steps −if matrix is empty, thenreturn −1first := a new set from first row of matrixfor each row in matrix, dofirst := Intersect first a set of elements of rowif first is empty, thenreturn −1return minimum of firstLet us see the following implementation to get better understanding −Example Live Democlass Solution: ... Read More

Find Sibling Value of a Binary Tree Node in Python

Arnab Chakraborty
Updated on 21-Oct-2020 11:03:46

667 Views

Suppose we have a value k and a binary search tree, here each node is either a leaf or contains 2 children. We have to find the node containing the value k, and return its sibling's value.So, if the input is likek = 4., then the output will be 10.To solve this, we will follow these steps −Define a function util() . This will take root, k, ansif left of root is not null and right of root is not null, thenreturnif k > value of root, thenif value of right of root is same as k, theninsert value of ... Read More

Split Set into Equal Sum Sets in Python

Arnab Chakraborty
Updated on 21-Oct-2020 11:00:47

856 Views

Suppose we have a list of numbers called nums, we have to check whether we can divide the list into two groups A and B such that: The sum of A and the sum of B are same. Here every number in A is strictly smaller than every number in B.So, if the input is like nums = [3, 4, 5, 12], then the output will be True, as we can have A = [3, 4, 5] and B = [12] and both have sum 12.To solve this, we will follow these steps −sort the list numstotal := sum of ... Read More

Count Total Number of Set Bits in Range 0 to N in Python

Arnab Chakraborty
Updated on 21-Oct-2020 10:58:45

221 Views

Suppose we have a number num. For each numbers i in the range 0 ≤ i ≤ num we have to calculate the number of 1's in their binary counterpart and return them as a list. So if the number is 5, then the numbers are [0, 1, 2, 3, 4, 5], and number of 1s in these numbers are [0, 1, 1, 2, 1, 2], so it will return 7.To solve this, we will follow these steps −res := an array which holds num + 1 number of 0soffset := 0for i in range 1 to num + 1if ... Read More

Count Minimum Number of Animals with No Predator in Python

Arnab Chakraborty
Updated on 21-Oct-2020 10:56:48

1K+ Views

Suppose we have a list of numbers called nums where nums[i] shows the predator of the ith animal and if there is no predator, it will hold −1. We have to find the smallest number of groups of animals such that no animal is in the same group with its direct or indirect predator.So, if the input is like nums = [1, 2, −1, 4, 5, −1], then the output will be 3, as we can have the groups like: [0, 3], [1, 4], [2, 5].To solve this, we will follow these steps −if A is empty, thenreturn 0adj := ... Read More

Separate Persons Where No Enemies Can Stay in Same Group in Python

Arnab Chakraborty
Updated on 21-Oct-2020 10:52:05

921 Views

Suppose we have a number n and a 2D matrix called enemies. Here n indicates there is n people labeled from [0, n - 1]. Now each row in enemies contains [a, b] which means that a and b are enemies. We have to check whether it is possible to partition the n people into two groups such that no two people that are enemies are in the same group.So, if the input is like n = 4, enemies = [[0, 3], [3, 2]], then the output will be True, as we can have these two groups [0, 1, 2] ... Read More

Reverse a Sentence with Words Stored as Character Array in C++

Arnab Chakraborty
Updated on 21-Oct-2020 10:39:52

1K+ Views

Suppose we have one input string sentence where each element is stored as single character, we have to reverse the strings word by word.So, if the input is like ["t", "h", "e", " ", "m", "a", "n", " ", "i", "s", " ", "n", "l", "c", "e"], then the output will be ["n", "l", "c", "e", " ", "i", "s", " ", "m", "a", "n", " ", "t", "h", "e"]To solve this, we will follow these steps −reverse the array sj := 0n := size of sfor initialize i := 0, when i < n, update (increase i by 1), ... Read More

Python Task Interval Finder

Arnab Chakraborty
Updated on 21-Oct-2020 10:36:40

126 Views

Suppose we have a list of intervals where each interval is like [start, end) and we also have a list of strings called types. Now for a given i, the intervals[i] shows the times someone worked on the job types[i] from [start, end). Two intervals of the same type never overlap or touch. So we have to find a sorted merged list where each item has [start, end, num_types], indicates from start to end, num_types number of tasks were being worked on.So, if the input is like intervals = [ [0, 3], [5, 7], [0, 7] ] types = ["problem ... Read More

Rotate a Linked List by K Places in C++

Arnab Chakraborty
Updated on 21-Oct-2020 10:34:18

365 Views

Suppose we have a linked list. We have to rotate the list to the right by k places. The value of k will be positive. So if the list is like [1 −> 2 −> 3 −> 4 −> 5 −> NULL], and k = 2, then the output will be [4 −> 5 −> 1 −> 2 −> 3 −> NULL]Let us see the steps −If the list is empty, then return nulllen := 1create one node called tail := headwhile next of tail is not nullincrease len by 1tail := next of tailnext of tail := headk := ... Read More

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