Ad
darkblacksidevs.ERS_B

Most frequently appearing number in given array

It is possible to exit before the end is reached, but in this example I added complexity to the model via the sorting.

Code
Diff
  • import java.util.Arrays;

class MaxOccurence {

  
  public static int findMax(int[] nums) {
      Arrays.sort(nums);
      int countNumberMax = -1;
      int numberMax = -1;
      
      int currentCount = 0;
      int currentNumber = nums.length>0?nums[0]:-1;
    
      for(int i = 0; i < nums.length; i++)
      {        
        if(currentNumber == nums[i])
        {
          currentCount++;
        
          if(countNumberMax < currentCount)
          {
            countNumberMax = currentCount;
            numberMax = currentNumber;
          }
        }
        else
        {
          currentCount = 1;
          currentNumber = nums[i];
        }
        
        if(countNumberMax > nums.length/2 || countNumberMax > currentCount + nums.length - i)
        {
          break;
        }
      }
      
      return numberMax;
  }
}
    • import java.util.Map;
    • import java.util.HashMap;
    • import java.util.Arrays;
    • class MaxOccurence {
    • public static int findMax(int[] nums) {
    • HashMap<Integer,Integer> map = new HashMap<>();
    • int lastBigVal = -1;
    • int lastBigValIndx = -1;
    • for(int i=0;i<nums.length;i++){
    • map.putIfAbsent(nums[i],0);
    • if(map.containsKey(nums[i])) {
    • map.put(nums[i], map.get(nums[i]) + 1);
    • if(lastBigVal < map.get(nums[i])) {
    • lastBigVal = map.get(nums[i]);
    • lastBigValIndx = nums[i];
    • }
    • }
    • }
    • return lastBigValIndx;
    • Arrays.sort(nums);
    • int countNumberMax = -1;
    • int numberMax = -1;
    • int currentCount = 0;
    • int currentNumber = nums.length>0?nums[0]:-1;
    • for(int i = 0; i < nums.length; i++)
    • {
    • if(currentNumber == nums[i])
    • {
    • currentCount++;
    • if(countNumberMax < currentCount)
    • {
    • countNumberMax = currentCount;
    • numberMax = currentNumber;
    • }
    • }
    • else
    • {
    • currentCount = 1;
    • currentNumber = nums[i];
    • }
    • if(countNumberMax > nums.length/2 || countNumberMax > currentCount + nums.length - i)
    • {
    • break;
    • }
    • }
    • return numberMax;
    • }
    • }