hashmap

Author: anishkumar127

Interview Ready Solution/Preparation 🔥/Other/hashmap/medium/Equality in Array.md

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+```java
+import java.util.*;
+
+class Main {
+  public static void main(String[] args) {
+      Scanner sc = new Scanner(System.in);
+      
+      int n = sc.nextInt();
+      int[] arr = new int[n];
+
+      for (int i=0;i < n;i++) {
+          arr[i] = sc.nextInt();
+      }
+
+      System.out.println(makeUnique(arr,n));
+      
+  }
+    static int makeUnique (int[] arr,int n) {
+     //Write your code here
+		HashMap<Integer,Integer> map  = new HashMap<>();
+
+		for(int val:arr){
+			map.put(val,map.getOrDefault(val,0)+1);
+		}
+
+		int max = Integer.MIN_VALUE;
+		for(Map.Entry<Integer,Integer>entry:map.entrySet()){
+			max = Math.max(max,entry.getValue());
+		}
+		return n-max;
+    }
+
+}
+
+
+```
+
+
+Equality in Array
+Given an array of integers, determine the minimum number of elements to delete to leave only elements of equal value.
+
+Input Format
+The first line contains an integer n, the number of elements in arr.
+
+The next line contains n space-separated integers arr[i].
+
+Output Format
+A single integer denoting the minimum number of deletions.
+
+Example 1
+Input
+
+4
+1 2 2 3
+Output
+
+2
+Explanation
+
+Delete the 2 elements 1 and 3 leaving arr as [2,2]. If both twos plus either the 1 or the 3 are deleted, it takes 3 deletions to leave either [3] or [1]. The minimum number of deletions is 2.
+
+Example 2
+Input
+
+5
+3 3 2 1 3
+Output
+
+2
+Explanation
+
+Delete 2 and 1 to leave [3,3,3]. This is minimal. The only other options are to delete 4 elements to get an array of either [1] or [2].
+
+Constraints
+0<n<=100
+
+1<=arr[i]<=100