nCr calculation code in DSA/Java

DSA/Java/Dynamic_Programming/nCr Calculation/Problem Statement

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+Find nCr for given n and r.
+
+Example 1:
+
+Input:
+n = 3, r = 2
+Output: 3
+Example 2:
+
+Input:
+n = 4, r = 2
+Output: 6
+Your Task:
+Complete the function nCrModp() which takes two integers n and r as input parameters and returns the nCr mod 109+7.
+Note: nCr does not exist when r > n. Hence, return 0 in that case.
+
+Expected Time Complexity : O(N*R)
+Expected Auxiliary Space: O(N)
+
+Constraints:
+1<= n <= 103
+1<= r <= 103
+
+Company Tags
+ Amazon

DSA/Java/Dynamic_Programming/nCr Calculation/Solution.java

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+class Solution
+{
+
+    static int p = 1000000007;
+
+    public static int min(int a, int b) 
+    { 
+        return (a<b)? a: b;  
+    }
+
+    //Function to return nCr mod 10^9+7 for given n and r.
+    public static int nCrModp(int n, int r)
+    {
+        //array C[] is going to store last row of pascal triangle at the end.
+        //last entry of last row is nCr. 
+        int C[] = new int[r+1];
+        Arrays.fill(C, 0);
+
+        //top row of Pascal Triangle 
+        C[0] = 1; 
+
+        //one by one constructing remaining rows of Pascal 
+        //triangle from top to bottom.    
+        for (int i = 1; i <= n; i++) 
+        { 
+          //filling entries of current row using previous row values. 
+          for (int j = min(i, r); j > 0; j--) 
+            // nCj = (n-1)Cj + (n-1)C(j-1); 
+            C[j] = (C[j]%p + C[j-1]%p)%p; 
+        } 
+
+        //returning the result.
+        return C[r];
+    }
+}