INTERN

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);

}