Subarrays with K Different Integers | Count Subarrays Where Max Element Appears at Least K Times

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sir very easy atmost code (thak you sir) class Solution { public: int subarraysWithKDistinct(vector& nums, int k) { //at least(k)-at least(k-1) int total=0; unordered_mapferq; // use count of different integer int start=0,end=0,count=0; int n=nums.size(); while(endk &&start

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class Solution { public: int atMostK(vector& nums, int k){ int start=0,end=0; int ans=0; unordered_mapm; int total = 0; int n=nums.size(); while(endk && start

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thank u sirji

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3rd Problem with Atmost :- class Solution { public: int subarraysWithKDistinct(vector& nums, int k) { int start = 0 , end = 0 , total = 0; int count = 0; int n = nums.size(); unordered_map freq; while(end < n){ freq[nums[end]]++; if(freq[nums[end]] == 1) count++; while(count > k && start k && start

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Day 196 ✅🔥

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Homework Done - int subarraysWithKDistinct(vector& nums, int k) { //Atmost(k) - Atmpst(k-1) unordered_map freq; int start, end, n, count, total; start = end = count = total = 0; n = nums.size(); while(end < n) { freq[nums[end]]++; if(freq[nums[end]] == 1) count++; while(count > k) { freq[nums[start]]--; if(freq[nums[start]] == 0) count--; start++; } total += end - start + 1; end++; } freq.clear(); start = end = count = 0; k--; while(end < n) { freq[nums[end]]++; if(freq[nums[end]] == 1) count++; while(count > k) { freq[nums[start]]--; if(freq[nums[start]] == 0) count--; start++; } total -= end - start + 1; end++; } return total; }

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