HackerEarth The Old Monk problem solution

In this HackerEarth The Old Monk problem solution Big Chandan is a dire lover of Biryani, especially Old Monk's Biryani. Today, he went over to have some of it. To his surprise, the waiter turns out be to be a coding geek and refuses to serves him unless Chandu solves his two- arrays problem, stated as:

Given two non-increasing array of integers A,B i.e A[i] >= A[i+1] and B[i] >= B[i+1] and for all i, 0 ≤ i < n-1.

The monkiness of two numbers is given by: M (A[i],B[j]) = j - i , if j >=i and B[j] >= A[i], or 0 otherwise.

Find the monkiness of the two arrays, that is given by: M (A,B)= max (M(A[i],B[j])) for 0≤ i, j< n-1.

HackerEarth The Old Monk problem solution.

tc = int(raw_input())
while tc>0:
    tc = tc - 1
    curMax = 0
    n = int(raw_input())
    x = map(int, raw_input().split())
    y = map(int, raw_input().split())
    for i in xrange(n):
        low, high, pos = 0, n - 1, -1
        while low<=high:
             mid = (low + high) / 2
             if y[mid] >= x[i]:
                 pos = mid
                 low = mid + 1
             else:
                 high = mid - 1
        curMax = max (curMax, pos - i)
    print curMax

Second solution

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    assert(1 <= t && t <= 50);
    while(t--)
    {
        int n,i;
        scanf("%d",&n);
        assert(1 <= n && n <= 100000);
        vector < long long > a(n),b(n);
        for(i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            if(i)
                assert(a[i] <= a[i-1]);
            assert(1 <= a[i] && a[i] <= (long long)1e12);
        }
        for(i=0;i<n;i++)
        {
            scanf("%lld",&b[i]);
            if(i)
                assert(b[i] <= b[i-1]);
            assert(1 <= b[i] && b[i] <= (long long)1e12);
        }
        int ans = 0;
        for(i=0;i<n;i++)
        {
            int l = 0 , h = n-1 , m;
            int p = -1;
            while(l <= h)
            {
                m = (l+h)/2;
                if(b[m] >= a[i])
                {
                    p = m;
                    l = m + 1;
                }
                else
                {
                    h = m - 1;
                }
            }
            ans = max(ans,p - i);
        }
        printf("%d\n",ans);
    }
    return 0;
}