SRM 715

algorithm

Topcoder SRM 715

Q1 : ImageCompression 

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public class ImageCompression {
    public String isPossible(String[] input, int k) {
        boolean status = false;
        for(int i = 0; i < input.length; i+=k) {
            for(int j = 0; j < input[0].length(); j+=k) {
                char start = input[i].charAt(j);
                for(int l = i; l < i + k; l++)
                    for(int m = j; m < j + k; m++)
                    if (input[l].charAt(m) != start) {
                        System.out.println(l);
                    System.out.println(m);
                    return "Impossible";  
                    }
              }                
        }
        return "Possible";
    }
}

Q2: MaximumRangeDiv2

This question should be solved by greedy algorithm. But I didn’t realize that and choose to use brute force. I failed at only one test cases

Q3)

Failed to came up with the solution, I reference an accept user Abni_Saini:

Code Ref: 

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#include <bits/stdc++.h>
using namespace std;
typedef long long int lld;
#define rep(i,a,n) for(long long int i = (a); i <= (n); ++i)
#define repI(i,a,n) for(int i = (a); i <= (n); ++i)
#define repD(i,a,n) for(long long int i = (a); i >= (n); --i)
#define repDI(i,a,n) for(int i = (a); i >= (n); --i)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
typedef pair<lld,lld> PA;
class InPrePost{
public:
    bool check(vector<int> A,vector<int> B){
        sort(A.begin(),A.end());
        sort(B.begin(),B.end());
        int n = A.size();
        if(n!=B.size()) return false;
        rep(i,0,n-1) if(A[i]!=B[i]) return false;
        return true;
    }
 
    bool tellMe(vector<string> S, vector<int> a1, vector<int> a2, vector<int> a3){
        // Check if there are elements not equal.
        if(!check(a1,a2) || !check(a1,a3)) return false;
        int n = a1.size();
        if(n<=1) return true;
        int root = a1[0];
        vector<int> a1l,a1r,a2l,a2r,a3l,a3r;
        int leftn = -1;
        rep(i,0,n-1){
            if(a2[i]==root){
                leftn = i;
            }
            else if(leftn==-1){
                a2l.pb(a2[i]);
            }
            else a2r.pb(a2[i]);
        }
        rep(i,1,n-1){
            if(i<=leftn) a1l.pb(a1[i]);
            else a1r.pb(a1[i]);
        }
        rep(i,0,n-2){
            if(i<leftn) a3l.pb(a3[i]);
            else a3r.pb(a3[i]);
        }
        map<string,vector<int> > M;
        M[S[0]] = a1l;
        M[S[2]] = a2l;
        M[S[4]] = a3l;
        if(!tellMe(S,M["pre"],M["in"],M["post"])) return false;
        M[S[1]] = a1r;
        M[S[3]] = a2r;
        M[S[5]] = a3r;
        if(!tellMe(S,M["pre"],M["in"],M["post"])) return false;
        return true;
    }
 
    string isPossible(vector<string> S, vector<int> a1, vector<int> a2, vector<int> a3){
        if(tellMe(S,a1,a2,a3)) return "Possible";
        else return "Impossible";
    }
};
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