## Solutions

We can check each element against its neighbors’ depth and mark it with ‘X’ if the cavity condition is satisfied.

Here is a C++ Solution:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27  size_t n; cin >> n; vector> data; data.reserve(n); for (auto i=0u; i(cin), n, begin(data[i])); } for (auto i=1u; i (data[i][j+1]-'0') && cavity > (data[i][j-1]-'0') && cavity > (data[i-1][j]-'0') && cavity > (data[i+1][j]-'0')) data[i][j] = 'X'; } } for (const auto& row : data) { copy(begin(row), end(row), ostream_iterator(cout)); cout << endl; } 

## Day of the Programmer

Solutions This problem seems convoluted but it can be solved very easily if decomposed in terms of simple predicates: if the year is 1918, we can directly return the day 26.09.1918 otherwise we can check if the year is leap. In this case we return 12.09 and the year if the year is not leap, we just return 13.09 and the year To determine if a year is leap, we have to apply either the Julian - if the year is less than 1918 - or the Gregorian rule - otherwise. [Read More]

## Fair Rations

You are the benevolent ruler of Rankhacker Castle, and today you’re distributing bread. Your subjects are in a line, and some of them already have some loaves. Times are hard and your castle’s food stocks are dwindling, so you must distribute as few loaves as possible according to the following rules: Every time you give a loaf of bread to some person , you must also give a loaf of bread to the person immediately in front of or behind them in the line (i. [Read More]

Given an array of integers where each element is obtained by adding either +1 or -1 to the previous element you need to find an element index with the minimum number of comparisons. If the element is present multiple times then print the smallest index. If the element is not present print -1. Input Format int N, denoting the number of elements. int T, denoting the number of queries. N space separated elements, denoting the array to search. [Read More]

## Hotel Coverage

Solutions This problem falls into the greedy category. A hotel can accomodate all customers in a distance of $$-k$$ and $$+k$$. We should find the smallest amount of intervals of length $$2k$$ that cover all the positions of the customers. First of all, we sort the positions of the customers. We iterate over all the sorted positions and, at each step, we calculate the difference between adjacent positions. We keep track of the interval by incrementing a running sum of the such differences. [Read More]

## New Year Chaos

Solutions We can see, intuitively, that the number of people a person $$p$$ has bribed corresponds to the number of people on her right with value less than $$p$$. For instance: 1 1 2 3 5 4 In this case, $$5$$ has bribed $$4$$ and took her position. Another example: 1 2 1 5 3 4 We have three bribes: $$2$$ has bribed $$1$$ $$5$$ has bribed $$4$$ $$5$$ has bribed $$3$$ This problem is equivalent to calculating the number of inversions of the array, that is the number of elements that are out of the natural order of a sequence. [Read More]

## One Swap

Bob has an array $$a$$ with $$n$$ positive integer elements. Bob likes order, so he wants his array to be sorted in non-decreasing order. He decides to swap two elements in the array to make his array sorted. (A swap is defined as switching two elements at distinct locations in the array.) Your task is to determine if this can be done. If he can’t sort the array with one swap, print $$-1$$. [Read More]