Create 20.1 Insertion Sort.java

for(int i=1;i<n;i++)
        {
            int key = Z[i];
            int j = i-1;
            while(j>=0)
            {
                if(key>Z[j])
                {
                    Z[j+1] = Z[j];
                    Z[j] = key;
                }
                j--;
            }
            
        }

Author: prokrishpatel

20.1 Insertion Sort.java

@@ -0,0 +1,65 @@
+/*
+Akash is found of collecting unique coins. Every coin has a denomination i.e Z. One Afternoon, he has K coins. He wnat to arrange K coins in non-increaasing order as per denomination. Write a programming solution to help akash. - Note: Use Insertion Sort
+
+Input Format
+
+1st line contain K Coins
+2nd line contain space separated denomination from K coins
+Constraints
+
+K>0 & K<50
+Z>0 & Z<100
+Output Format
+
+space separated sorted denomination from K coins
+
+Sample Input 0
+
+5
+10 4 3 7 8
+Sample Output 0
+
+10 8 7 4 3
+*/
+
+import java.io.*;
+import java.util.*;
+import java.text.*;
+import java.math.*;
+import java.util.regex.*;
+
+public class Solution {
+    public static void main(String args[] ) throws Exception {
+        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
+        Scanner sc = new Scanner(System.in);
+        //take input for number of coins: n
+        int n = sc.nextInt();
+        //take input for domination of each coin ie. Z as an array
+        int [] Z = new int [n];
+        for(int i=0;i<n;i++)
+            Z[i] = sc.nextInt();
+        
+        //for sorting we use INSERTION SORT 
+        
+        for(int i=1;i<n;i++)
+        {
+            int key = Z[i];
+            int j = i-1;
+            while(j>=0)
+            {
+                if(key>Z[j])
+                {
+                    Z[j+1] = Z[j];
+                    Z[j] = key;
+                }
+                j--;
+            }
+            
+        }
+        
+        //printing the final sorted array ie. Z
+        
+        for(int i:Z)
+            System.out.print(i+" ");
+    }
+}