Intersection of two numerical arrays
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You are given a fixed number of 5 rupee, 10 rupee, 20 rupee and 50 rupee stamps. Now given an amount for sending a parcel, you should design an algorithm to come out with the minimum number of stamps that should be used for attaining that amount. For example, if the parcel costed 30 rupees, it could be attained using one 20 rupee stamp and one 10 rupee stamp OR using three 10 rupee stamps OR using one 20 rupee stamp and two 5 rupee stamps OR using one 10 rupee stamp and four 5 rupee stamps OR using two 10 rupee stamps and two 5 rupee stamps. However, the minimum number of stamps is the case of one 20 rupee stamp and one 10 rupee stamp where only two stamps are used. The case where no solution is possible should also be handled, for example, a parcel amount of exactly 33 rupees cannot be attained.
if there are 6 people in a team, how many handshakes will be there
Given a set of people, one of them is a celebrity. You have a 2D array which describes which people know each other, that is [N, M] is true if N knows M. The celebrity will not know anyone (except them self) and everyone will know the celebrity. Find an order N algorithm to find the celebrity.
1. Given a array of integers find the index which partitions the array to two with high numbers and low numbers. For example [5, -1, 3, 8,6] the index 3 will partition the array to [5,-1,3] and [8,6] all the numbers in the second partition are greater than first. The solution has to work in O(n).
Give an array that has only the values 1, 2 or 3, sort the array in place.
Given an array of numbers, replace each number with the product of all the numbers in the array except the number itself *without* using division.
Write a C function to return the number of set bits in an integer.
Explain the data structure you would use to implement pop() and push(Object, int) for a Priority Queue.
what's wrong with the following code : <template type T > T accumulate ( vector<T> in) { T total = in[0]; for (int i =0; i < in.length() ; i++) { total = total + in[i]; } return T }
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