Project Engineer

I initially went on to make 2 nested loops and for each iteration , I checked if arr[j]==arr[i] , I increased the count of arr[i] . At the end of inner loop , I compared the ans with this count and updated the ans with this count if this count>ans.

However , I then optimised this using a hashmap. I stored the frequencies & then traversed the map & checked , if current frequency is greater than largest stored frequency , then this can be the largest frequency element , & earlier stored frequency can be the second largest frequency. So make this as maxFreq & take the earlier stored maxFreq & make it as secondMax , if any other frequency occurs where that freqsecondMax , then , this freq can be the second Max freq element. Thus , I solved it in O(N) time.

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