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Google Kick Round-D Question (2020)

  • Difficulty Level : Easy
  • Last Updated : 04 Aug, 2021

Isyana is given the number of visitors at her local theme park on N consecutive days. The number of visitors on the i-th day is VI. A day is record-breaking if it satisfies both of the following conditions:

  1. The number of visitors on the day is strictly larger than the number of visitors on each of the previous days.
  2. Either it is the last day, or the number of visitors on the day is strictly larger than the number of visitors on the following day.

Note that the very first day could be a record-breaking day. Please help Isyana find out the number of record-breaking days.

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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 integer N. The second line contains N integers. The i-th integer is Vi.



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 record-breaking days.

Limits

Time limit: 20 seconds per test set.

Memory limit: 1GB.

1 ≤ T ≤ 100.
0 ≤ Vi ≤ 2 × 105.
Test set 1
1 ≤ N ≤ 1000.
Test set 2
1 ≤ N ≤ 2 × 105 for at most 10 test cases.
For the remaining cases, 1 ≤ N ≤ 1000.

Sample

Input
Output
4
8
1 2 0 7 2 0 2 0
6
4 8 15 16 23 42
9
3 1 4 1 5 9 2 6 5
6
9 9 9 9 9 9
Case #1: 2
Case #2: 1
Case #3: 3
Case #4: 0

In Sample Case #1: The bold and underlined numbers in the following represent the record-breaking days: 1 2 0 7 2 0 2 0.

In Sample Case #2: only the last day is a record-breaking day.

In Sample Case #3: The first, the third, and the sixth days are record-breaking days.

In Sample Case #4: there is no record-breaking day.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int calculate(vector<int>& arr, int n)
{
    int cnt = 0;
    // initialising prevmax as -infinity
    int prevmax = INT_MIN;
 
    for (int i = 0; i < n; i++) {
        // check for condition 2
        if (i == n - 1) {
            if (arr[i] > prevmax)
                cnt++;
        }
        // check for condition 1
        else if (arr[i] > prevmax && arr[i] > arr[i + 1]) {
            cnt++;
        }
        // update prevmax
        prevmax = max(prevmax, arr[i]);
    }
 
    return cnt;
}
 
int main()
{
 
    int t = 1;
    // Taking the input for every test case
    while (t--) {
        int n = 9;
        vector<int> arr = { 3, 1, 4, 1, 5, 9, 2, 6, 5 };
        cout << calculate(arr, n) << endl;
    }
 
    return 0;
}

Java




// JAva program for the above approach
import java.io.*;
 
class GFG{
 
    static int calculate(int[] arr, int n)
    {
        int cnt = 0;
        // initialising prevmax as -infinity
        int prevmax = Integer.MIN_VALUE;
     
        for (int i = 0; i < n; i++) {
            // check for condition 2
            if (i == n - 1) {
                if (arr[i] > prevmax)
                    cnt++;
            }
            // check for condition 1
            else if (arr[i] > prevmax && arr[i] > arr[i + 1]) {
                cnt++;
            }
            // update prevmax
            prevmax = Math.max(prevmax, arr[i]);
        }
     
        return cnt;
    }
 
    // Driver Code
    public static void main (String[] args)
    {
     
        int t = 1;
        // Taking the input for every test case
        while (t-- != 0) {
            int n = 9;
            int arr[] = { 3, 1, 4, 1, 5, 9, 2, 6, 5 };
            System.out.println(calculate(arr, n));
        }
    }
     
}
 
// This code is contributed by shubhamsingh10

Python3




# Python program for the above approach
def calculate(arr, n):
     
    cnt = 0
     
    # initialising prevmax as -infinity
    prevmax = -2**31
     
    for i in range (n):
       
        # check for condition 2
        if (i == n - 1):
            if (arr[i] > prevmax):
                cnt+=1
         
        # check for condition 1
        elif(arr[i] > prevmax and arr[i] > arr[i + 1]):
            cnt+=1
         
        # update prevmax
        prevmax = max(prevmax, arr[i])
    return cnt
     
t = 1
# Taking the input for every test case
while (t) :
    n = 9
    arr =  [3, 1, 4, 1, 5, 9, 2, 6, 5]
    print(calculate(arr, n))
    t-=1
 
    # This code is contributed by shivanisinghss2110.

C#




// C# program for the above approach
using System;
 
public class GFG{
     
    static int calculate(int[] arr, int n)
    {
        int cnt = 0;
       
        // initialising prevmax as -infinity
        int prevmax = Int32.MinValue;
     
        for (int i = 0; i < n; i++)
        {
           
            // check for condition 2
            if (i == n - 1) {
                if (arr[i] > prevmax)
                    cnt++;
            }
           
            // check for condition 1
            else if (arr[i] > prevmax && arr[i] > arr[i + 1]) {
                cnt++;
            }
            // update prevmax
            prevmax = Math.Max(prevmax, arr[i]);
        }
     
        return cnt;
    }
 
    // Driver Code
    static public void Main ()
    {
        int t = 1;
       
        // Taking the input for every test case
        while (t-- != 0)
        {
            int n = 9;
            int[] arr = { 3, 1, 4, 1, 5, 9, 2, 6, 5 };
            Console.Write(calculate(arr, n));
        }
    }
}
 
// This code is contributed by shubhamsingh10

Javascript




<script>
 
// Javascript program for the above approach
function calculate(arr, n)
{
    var cnt = 0;
     
    // Initialising prevmax as -infinity
    var prevmax = Number.MIN_VALUE;
 
    for(var i = 0; i < n; i++)
    {
         
        // Check for condition 2
        if (i == n - 1)
        {
            if (arr[i] > prevmax)
                cnt++;
        }
         
        // Check for condition 1
        else if (arr[i] > prevmax &&
                 arr[i] > arr[i + 1])
        {
            cnt++;
        }
         
        // Update prevmax
        prevmax = Math.max(prevmax, arr[i]);
    }
    return cnt;
}
 
// Driver Code
var t = 1;
 
// Taking the input for every test case
while (t-- != 0)
{
    var n = 9;
    var arr = [ 3, 1, 4, 1, 5, 9, 2, 6, 5 ];
     
    document.write(calculate(arr, n));
}
 
// This code is contributed by shivanisinghss2110
 
</script>
Output
3



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