Google Kick start Round C 2020
Problem
Avery has an array of N positive integers. The i-th integer of the array is Ai.
A contiguous subarray is an m-countdown if it is of length m and contains the integers m, m-1, m-2, …, 2, 1 in that order. For example, [3, 2, 1]
is a 3-countdown.
Can you help Avery count the number of K-countdowns in her array?
Solution:
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the integers N and K. The second line contains N integers. The i-th integer is Ai.
Output
For each test case, output one line containing Case #x: y
, where x
is the test case number (starting from 1) and y
is the number of K-countdowns in her array.
Limits
Time limit: 20 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
2 ≤ K ≤ N.
1 ≤ Ai ≤ 2 × 105, for all i.
Test set 1
2 ≤ N ≤ 1000.
Test set 2
2 ≤ N ≤ 2 × 105 for at most 10 test cases.
For the remaining cases, 2 ≤ N ≤ 1000.
Sample
Input
Output
3
12 3
1 2 3 7 9 3 2 1 8 3 2 1
4 2
101 100 99 98
9 6
100 7 6 5 4 3 2 1 100Case #1: 2
Case #2: 0
Case #3: 1
In sample case #1, there are two 3-countdowns as highlighted below.
- 1 2 3 7 9 3 2 1 8 3 2 1
- 1 2 3 7 9 3 2 1 8 3 2 1
In sample case #2, there are no 2-countdowns.
In sample case #3, there is one 6-countdown as highlighted below.
- 100 7 6 5 4 3 2 1 100
Code:
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
int n,k;
for(int i=0;i<t;i++){
cin>>n>>k;
int ans=0;
int arr[n];
for(int j=0;j<n;j++){
cin>>arr[j];
}
for(int j=0;j<n;j++){
if(arr[j]==k){
int x=k;
for(int a=j;a<j+k;a++){
if(arr[a]==x){
x — ;
}
else{
break;
}
}
if(x==0){
ans++;
}
}
}
cout<<”Case #”<<i+1<<”: “<<ans<<endl;
}
}
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