Created
October 23, 2021 16:12
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How Many Numbers Are Smaller Than the Current Number
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| class Solution { | |
| public int[] smallerNumbersThanCurrent(int[] nums) { | |
| int[] ans = new int[nums.length]; | |
| // Brute Technique | |
| // for(int i=0;i<nums.length;i++) | |
| // { | |
| // int c=0; | |
| // for(int j=0;j<nums.length;j++) | |
| // { | |
| // if(nums[i]>=nums[j] && nums[i] != nums[j]) | |
| // { | |
| // c++; | |
| // } | |
| // } | |
| // ans[i] = c; | |
| // } | |
| // Another Approach | |
| int[] helper = new int[nums.length]; | |
| for(int i=0;i<nums.length;i++) | |
| { | |
| helper[i] = nums[i]; | |
| } | |
| // for(int i=0;i<helper.length;i++) | |
| // { | |
| // for(int j=i+1;j<helper.length;j++) | |
| // { | |
| // if(helper[i]>=helper[j]) | |
| // { | |
| // int t = helper[i]; | |
| // helper[i] = helper[j]; | |
| // helper[j] = helper[i]; | |
| // } | |
| // } | |
| // } | |
| Arrays.sort(helper); | |
| HashMap<Integer,Integer> hm = new HashMap<>(); | |
| hm.put(helper[0],0); | |
| for(int i=1;i<helper.length;i++) | |
| { | |
| if(helper[i] == helper[i-1]) | |
| { | |
| hm.put(helper[i],hm.get(helper[i-1])); | |
| } | |
| else{ | |
| hm.put(helper[i],i); | |
| } | |
| } | |
| for(int i=0;i<nums.length;i++) | |
| { | |
| if(hm.containsKey(nums[i])) | |
| ans[i] = hm.get(nums[i]); | |
| } | |
| return ans; | |
| } | |
| } |
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