Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Android Unlock Patterns in C++
Suppose we have an Android 3x3 key lock screen and two integers m and n, the values of m and n are in range 1 ≤ m ≤ n ≤ 9, We have to count the total number of unlock patterns of the Android lock screen, which consist of minimum of m keys and maximum n keys.
The rule is like, each pattern must connect at least m keys and at most n keys. All the keys must be unique. If there is a line connecting two consecutive keys in the pattern passes through any other keys, the other keys must have previously selected in the pattern. Jumping through any key that is not selected that is not allowed. The order of keys used matters.
So, if the input is like m = 1, n = 1, then the output will be 9
To solve this, we will follow these steps −
Define an array skip of size: 10 x 10.
Define a function dfs(), this will take node, len, an array visited,
if len is same as 0, then −
return 1
visited[node] := true
ret := 0
for initialize i := 1, when i <= 9, update (increase i by 1), do −
if visited[i] is false and (skip[node, i] is same as 0 or visited[skip[node, i]] is nonzero), then −
ret := ret + dfs(i, len - 1, visited)
visited[node] := false
return ret
From the main method do the following −
fill skip with 0
skip[1, 3] := skip[3, 1] := 2
skip[1, 7] := skip[7, 1] := 4
skip[3, 9] := skip[9, 3] := 6
skip[7, 9] := skip[9, 7] := 8
skip[4, 6] := skip[6, 4] := skip[2, 8] := skip[8, 2] := skip[3, 7] := skip[7, 3] := skip[1, 9] := skip[9, 1] := 5
Define an array visited of size 10
ret := 0
for initialize i := m, when i <= n, update (increase i by 1), do −
ret := ret + (dfs(1, i - 1, visited))
ret := ret + (dfs(2, i - 1, visited))
ret := ret + dfs(5, i - 1, visited)
return ret
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int skip[10][10];
int dfs(int node, int len, vector<bool>& visited){
if (len == 0)
return 1;
visited[node] = true;
int ret = 0;
for (int i = 1; i <= 9; i++) {
if (!visited[i] && (skip[node][i] == 0 || visited[skip[node][i]])) {
ret += dfs(i, len - 1, visited);
}
}
visited[node] = false;
return ret;
}
int numberOfPatterns(int m, int n){
memset(skip, 0, sizeof(skip));
skip[1][3] = skip[3][1] = 2;
skip[1][7] = skip[7][1] = 4;
skip[3][9] = skip[9][3] = 6;
skip[7][9] = skip[9][7] = 8;
skip[4][6] = skip[6][4] = skip[2][8] = skip[8][2] = skip[3][7] = skip[7][3] = skip[1][9] = skip[9][1] = 5;
vector<bool> visited(10);
int ret = 0;
for (int i = m; i <= n; i++) {
ret += (dfs(1, i - 1, visited) * 4);
ret += (dfs(2, i - 1, visited) * 4);
ret += dfs(5, i - 1, visited);
}
return ret;
}
};
main(){
Solution ob;
cout << (ob.numberOfPatterns(1,1));
}Input
1, 1
Output
9
809 Views
