Functional Array FUNARR Solution

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Functional Array FUNARR Solution

The functional array of an array A=[A1,A2,…,AN] is the array fA of size N−1, where fAi=Ai+1−Ai for 1≤i<N. For example, if A=[2,3,9,11] then fA=[1,6,2].

You are given two arrays B=[B1,B2,…,BN] and Z=[Z1,Z2,…,ZM] of length N and M respectively. Find out whether there exists an array A such that:

  • B is a subsequence of A, and
  • fA is a subsequence of Z

Print “YES” (without quotes) if such an array A exists, and “NO” otherwise.

Input Format

  • The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follows.
  • Each test case contains three lines of input.
  • The first line contains two integers N and M the lengths of B and Z respectively.
  • The second line contains N space-separated integers B1,B2,…,BN; the elements of B.
  • The third line contains M space-separated integers Z1,Z2,…,ZM; the elements of Z.

Output Format

For each test case, output a single line containing one word – either “YES” or “NO”. You may print each character of the string in uppercase or lowercase (for example, the strings “yEs”, “yes”, “Yes” and “YES” will all be treated as identical).

Constraints

  • 1≤T≤105
  • 1≤N≤105
  • 1≤M≤105
  • 0≤Bi≤105
  • 0≤Zi≤105
  • The sum of N over all test cases does not exceed 105
  • The sum of M over all test cases does not exceed 105

Subtasks

Subtask #1 (5 points):

  • 1≤M≤10
  • 1≤N≤103
  • 0≤Bi≤103
  • 0≤Zi≤103
  • The sum of M over all test cases does not exceed 100

Subtask #2 (40 points):

  • 1≤M≤103
  • 1≤N≤103
  • 0≤Bi≤103
  • 0≤Zi≤103
  • The sum of M over all test cases does not exceed 103

Subtask #3 (55 points):

  • Original constraints

Sample Input 1

2
2 2
2 3
1 1
2 2
2 4
1 3

Sample Output 1

YES
NO

Explanation

Test case 1:

A=[1,2,3] is one possible array.

  • B=[2,3] is a subsequence of A
  • fA=[1,1] is a subsequence of Z=[1,1].

Test case 2:

It can be shown that no array A can both contain B=[2,4] as a subsequence and also have fA be a subsequence of Z=[1,3]

SOLUTION

Program Python: Functional Array FUNARR Solution in Python

from collections import defaultdict, deque
from bisect import bisect_left
from sys import stdin
def subset_sum (A, k):
    subsets = defaultdict(list)
    subsets[0] = []
    
    for i in range(len(A)):
        old_subsets = subsets
        subsets = defaultdict()
        
        for (prev_sum, subset) in old_subsets.items():
            subsets[prev_sum] = subset
            new_sum = prev_sum + A[i]
            new_subset = subset + [i]
            if k == new_sum:
                new_subset.sort()
                return new_subset
            else:
                subsets[new_sum] = new_subset
    
    return [-1]
def solve():
    
    results = []
    
    for _ in range(int(stdin.readline().strip())):
        N, M = map(int, stdin.readline().strip().split())
        B = list(map(int, stdin.readline().strip().split()))
        Z = list(map(int, stdin.readline().strip().split()))
        
        indices = defaultdict(deque)
        
        if N == 1:
            results.append("YES")
            continue
        
        D = [B[i] - B[i-1] for i in range(1,N)]
        
        if N-1 > M:
            results.append("NO")
            continue
        if min(D) < min(Z):
            results.append("NO")
            continue
        
        for i in range(M):
            indices[Z[i]].append(i)
        
        x = -1
        
        for i in range(len(D)):
            d = D[i]
            ans = "NO"
            y1, y2 = 10**10, 10**10
            popcnt = 0
            
            if d in indices and len(indices[d]):
                y1 = indices[d][0]+1
                popcnt += 1
                
                while popcnt < len(indices[d]) and y1 <= x:
                    y1 = indices[d][popcnt]+1
                    popcnt += 1
                
                if y1 <= x:
                    y1 = 10**10
            s = 0 if x == -1 else x
            e = y1 if y1 != 10**10 else M-(len(D)-i-1)
            
            sub = subset_sum(Z[s:e], d)
            
            if sub[0] != -1:
                y2 = sub[-1]+s+1
                    
            x = min(y1, y2)            
            while len(indices[d]) and x > indices[d][0]:
                indices[d].popleft()
            if len(D)-i-1 > M-x:              
                break
            
            if x != 10**10:
                ans = "YES"
            else:
                break
        results.append(ans)
        
    print ('\n'.join(map(str, results)))
if __name__ == "__main__":
    solve()

Program C++: Functional Array FUNARR Solution in C++

#include <bits/stdc++.h>
using namespace std;
#define     ff                  first
#define     pb                  push_back
#define     ss                  second
#define     IOS                 ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define     all(v)              v.begin(), v.end()
#define     allr(v)             v.rbegin(), v.rend()
#define       int               long long
#define       endl             '\n'
const   int N = 3e5+5, Mod = 1e9 + 7;
int32_t main(){
        IOS;
        #ifndef ONLINE_JUDGE
        freopen("inputf.in", "r", stdin);
        //    freopen("outputf.in", "w", stdout);
        #endif 
        int t; cin >> t;
        while(t--) {
            int n, m; cin >> n >> m;
            int a[n], b[m];
            for(int i = 0; i < n; i++)
                cin >> a[i];
            for(int i = 0; i < m; i++){
                int x; cin >> x;
                b[i] = x;
            }
            int j = 0;
            int all = 0;
            int ansr = 1;
            for(int i = 1; i < n; i++){
                int l = a[i - 1];
                int flag = 0;
                int r = a[i];
                if(r - l == 0){
                    while(j < m){
                        if(b[j] == 0){
                            flag = 1;
                            j++;
                            break;
                        }
                        j++;
                    }
                }
                else {
                    int diff = r - l;
                    int sum = 0;
                    set<int> st;
                    st.insert(0);
                    while(j < m){
                        int rr = b[j];
                        sum += rr;
                        j++;
                        set<int> st2;
                        int k = 50;
                        auto it = st.end();
                        it--;
                        while(k--){
                            if(rr + *it <= diff){
                                st2.insert(rr + *it);
                            }
                            st2.insert(*it);
                            if(rr + *it == diff){
                                flag = 1;
                                break;
                            }
                            if(it == st.begin())
                                break;
                            it--;
                        }
                        if(flag == 1)
                            break;
                        st = st2;
                    }
                }
                if(flag == 0){
                    ansr = 0;
                    break;
                }
            }
            if(ansr)
                cout << "YES\n";
            else cout << "NO\n";
        }
        return 0;
}

November Long Challenge 2021 Solution

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