Colorful Strings

Author: spaceman-dev

Some_More_Strings/ColorfulStrings.cpp

@@ -0,0 +1,42 @@
+// { Driver Code Starts
+// Initial Template for C++
+#include <bits/stdc++.h>
+using namespace std;
+
+ // } Driver Code Ends
+// User function Template for C++
+class Solution {
+ public:
+ long long possibleStrings(int n, int r, int b, int g) 
+ {
+ // code here
+ long long fact[21];
+ fact[0] = 1;
+ for(int i = 1 ; i <= 20 ; i++)
+ fact[i] = fact[i-1] * i;
+ long long res = 0, sum = n - (r + b + g);
+ for(int i = 0 ; i <= sum ; i++)
+ {
+ for(int j = 0 ; j <= sum ; j++)
+ {
+ if(sum - (i + j) >= 0)
+ res += fact[n] / (fact[r + i] * fact[g + j] * fact[sum - i - j + b]);
+ }
+ }
+ return res;
+ }
+};
+
+// { Driver Code Starts.
+int main() {
+ int t;
+ cin >> t;
+ while (t--) {
+ int n, r, g, b;
+ cin >> n >> r >> g >> b;
+ Solution ob;
+ cout << ob.possibleStrings(n, r, b, g) << endl;
+ }
+ return 0;
+}
+ // } Driver Code Ends

Some_More_Strings/README.md

@@ -97,3 +97,15 @@
  Input: s1 = aacdb, s2 = gafd
  Output: cbgf
  Explanation: The common characters of s1 and s2 are: a, d. The uncommon characters of s1 and s2 are c, b, g and f. Thus the modified string with uncommon characters concatenated is cbgf.
+
+## 12. Colorful Strings: 
+ Find the count of all possible strings of size n.Each character of the string is either ‘R’, ‘B’ or ‘G’. In the final string there needs to be at least r number of ‘R’, at least b number of ‘B’ and at least g number of ‘G’ (such that r + g + b <= n). 
+ Example 1:
+ Input: n = 4, r = 1, g = 1, b = 1
+ Output: 36 
+ Explanation: No. of 'R' >= 1, 
+  No. of ‘G’ >= 1, No. of ‘B’ >= 1 and (No. of ‘R’) + (No. of ‘B’) + (No. of ‘G’) = n then following cases are possible: 
+  1. RBGR and its 12 permutation 
+  2. RBGB and its 12 permutation 
+  3. RBGG and its 12 permutation 
+ Hence answer is 36.