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# Google Software Engineering Intern, Fall 2019 – North America

• Difficulty Level : Easy
• Last Updated : 03 Jul, 2019

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=-1, A=7, A=0, A=7, A=-8. The function should return 2.

```Input : [-1, 7, 0, 7, -8]
Output : 2
```

 `#include ``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;``    ``a = -1, a = 7, a = 0, a = 7, a = -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 ``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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