Leetcode-Solutions/findingfloor.py at main · hawkh/Leetcode-Solutions · GitHub

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
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
# Finding the Floor
# Given an array, you have to find the floor of a number x. The floor of a number x is nothing but the largest number in the array less than or equal to x.


# Input Format

# The first line of input contains the N - the size of the array. The next line contains N integers, the elements of the array. The next line contains Q - number of queries. Each of the next Q lines contains a single integer X, for which you have to find the floor of X in the given array.


# Output Format

# For each query, print the floor of X, separated by a new line. If the floor is not found, print the value of "INT_MIN".


# Constraints

# 30 points

# 1 <= N <= 105

# 1 <= Q <= 102

# -108 <= ar[i] <= 108


# 70 points

# 1 <= N <= 105

# 1 <= Q <= 105

# -108 <= ar[i] <= 108


# Example

# Input

# 6

# -6 10 -1 20 15 5

# 5

# -1

# 10

# 8

# -10

# -4


# Output

# -1

# 10

# 5

# -2147483648

# -6


# Explanation


# Self Explanatory

n=int(input())
ar=list(map(int,input().split()))
ar.sort()
T=int(input())
for i in range(T):
    q=int(input())
    f=-2147483648
    l,h=0,n-1
    while(l<=h):
        mid=l+(h-l)//2
        if(ar[mid]==q):
            f=q
            break
        elif(ar[mid]>q):
            h=mid-1
        else:
            f=ar[mid]
            l=mid+1
    print(f)