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class Solution { public int compress(char[] chars) { int left =0; int right = 0; int k=0; int n= chars.length; while(right<n){ while(right<n && chars[right]==chars[left] ) { right++; } chars[k++]=chars[left]; int num= right-left; if (num > 1) { String countStr = Integer.toString(num); for (char c : countStr.toCharArray()) { chars[k++] = c; } } left=right; } return k; } }

class Solution { public boolean increasingTriplet(int[] nums) { if(nums.length<3) return false; int[] leftMin=new int[nums.length]; int[] rightMax=new int[nums.length]; int min=nums[0]; for(int i=0;i<nums.length;i++){ min=Math.min(nums[i],min); leftMin[i]=min; } int max=nums[nums.length-1]; for(int i=nums.length-1;i>=0;i--){ max=Math.max(nums[i],max); rightMax[i]=max; } for(int i=0;i<nums.length;i++){ if(leftMin[i]<nums[i] && nums[i]<rightMax[i]) return true; } return false; } }

class Solution { public String mergeAlternately(String word1, String word2) { int i=0,j=0; StringBuilder st= new StringBuilder(); while(i<word1.length() && j<word2.length()){ st.append(word1.charAt(i++)); st.append(word2.charAt(j++)); } while(i<word1.length()){ st.append(word1.charAt(i++)); } while(j<word2.length()){ st.append(word2.charAt(j++)); } return st.toString(); } }

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Nice Question !! class Solution { public boolean closeStrings(String word1, String word2) { if(word1.length()!=word2.length()) return false; int[] f1=new int[26]; int[] f2=new int[26]; for(int i=0;i<word1.length();i++){ f1[word1.charAt(i)-'a']++; } for(int i=0;i<word2.length();i++){ f2[word2.charAt(i)-'a']++; } // need to check that char in f1 and f2 must match for(int i=0;i<26;i++){ if(f1[i]==f2[i]) continue; if(f1[i]!=0 && f2[i]==0) return false; if(f1[i]==0 && f2[i]!=0) return false; } Arrays.sort(f1); Arrays.sort(f2); for(int i=0;i<26;i++){ if(f1[i]!=f2[i]) return false; } return true; } }

Keep it up buddy!! 👍 I am following you for blind 75 problems 👍

Hi, can you explain why freq[n] is += 1? Having the equal sign would not bar whether more than 1 digit/string was to occur more than once. But freq[n] += 0 wouldn't really be logical. I really appreciate this. I'm just starting my sophmore year so still a ways to go, but I've heard of the LeetCode a lot.

Appreciated 👍👍. More success to you mate.

Thanks for providing such valuable content. Your teaching style makes learning easier and more interesting.

The way you break down complex concepts is amazing. This video was exactly what I needed.

Thanks

i tried to read this topic from multiple teachers but ended up here . best video🙌🙌🙌