Google Software Engineering Intern, Fall 2019 – North America
1. Consider a binary tree of N vertices such that children of node k are 2*k and 2*k+1. Vertex 1 is the root of the tree and each node has an integer value associated with it.
Such a tree may be represented as an array of N integers by writing down values from consecutive nodes.
The tree can be represented as an array [-1, 7, 0, 7, -8].
A node is said to be at level x if the length of the shortest path between that node and root x-1. So, the root is at level 1, the children of root are at level 2, and so on.
Your task is to find the smallest level number x such that sum of all nodes at level x is maximal.
Examples: Given array A such that: A[0]=-1, A[1]=7, A[2]=0, A[3]=7, A[4]=-8. The function should return 2.
Input : [-1, 7, 0, 7, -8] Output : 2
#include <iostream>using namespace std;int solution(int a[], int n){ int max = -1; int temp = 0; for (int i = 0; i < n; i = i + 2) { if (i == 0) temp = a[i]; else temp = a[i] + a[i - 1]; if (temp > max) max = i; } return max;} int main(){ int a[4]; a[0] = -1, a[1] = 7, a[2] = 0, a[3] = 7, a[4] = -8; int size = 4; cout << solution(a, size);} |
2. Imagine you have a special keyboard with all keys in a single row. The layout of characters on a keyboard is denoted by a string S1 of length 26. S1 is indexed from 0 to 25. Initially, your finger is at index 0. To type a character, you have to move your finger to the index of the desired character. The time taken to move your finger from index i to index j is |j-i|, where || denotes absolute value.
Write a function solution(), that given a string S1 that describes the keyboard layout and a string S2, returns an integer denoting the time taken to type string S2.
Examples:
S1 = abcdefghijklmnopqrstuvwxyz
S2 = cba
Input : S1 = abcdefghijklmnopqrstuvwxyz, S2 = cba Output : 4
#include <bits/stdc++.h>using namespace std; int solution(string& s1, string& s2){ map<char, int> dict; for (int i = 0; i < 26; i++) { dict[s1[i]] = i; } int ans = 0; int prev = 0; for (int i = 0; i < s2.length(); i++) { ans = ans + abs(dict[s2[i]] - prev); prev = dict[s2[i]]; } return ans;} int main(){ string s1 = "abcdefghijklmnopqrstuvwxyz"; string s2 = "cba"; cout << solution(s1, s2);} |
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