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A left rotation operation on an array shifts each of the array's elements unit to the left. For example, if left rotations are performed on array , then the array would become .
Given an array of integers and a number, , perform left rotations on the array. Return the updated array to be printed as a single line of space-separated integers.
Function Description
Complete the function rotLeft in the editor below. It should return the resulting array of integers.
rotLeft has the following parameter(s):
- An array of integers .
- An integer , the number of rotations.
Input Format
The first line contains two space-separated integers and , the size of and the number of left rotations you must perform.
The second line contains space-separated integers .
Constraints
Output Format
Print a single line of space-separated integers denoting the final state of the array after performing left rotations.
Sample Input
5 4
1 2 3 4 5
Sample Output
5 1 2 3 4
Explanation
When we perform left rotations, the array undergoes the following sequence of changes:
#include <bits/stdc++.h>
using namespace std;
vector split_string(string);
// Complete the rotLeft function below.
vector rotLeft(vector a, int d) {
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
string nd_temp;
getline(cin, nd_temp);
vector nd = split_string(nd_temp);
int n = stoi(nd[0]);
int d = stoi(nd[1]);
string a_temp_temp;
getline(cin, a_temp_temp);
vector a_temp = split_string(a_temp_temp);
vector a(n);
for (int i = 0; i < n; i++) {
int a_item = stoi(a_temp[i]);
a[i] = a_item;
}
vector result = rotLeft(a, d);
for (int i = 0; i < result.size(); i++) {
fout << result[i];
if (i != result.size() - 1) {
fout << " ";
}
}
fout << "\n";
fout.close();
return 0;
}
vector split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});
input_string.erase(new_end, input_string.end());
while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}
vector splits;
char delimiter = ' ';
size_t i = 0;
size_t pos = input_string.find(delimiter);
while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));
i = pos + 1;
pos = input_string.find(delimiter, i);
}
splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
return splits;
}
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Sample Input
5 4
1 2 3 4 5
Sample Output
5 1 2 3 4