leetcode weekly 318

Author: Pankajcoder1

LeetCode/2460.Apply_Operations_to_an_Array.cpp

@@ -0,0 +1,58 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+
+class Solution
+{
+public:
+    vector<int> applyOperations(vector<int> &nums)
+    {
+        int n=nums.size();
+        vector<int> ans(n,0);
+        int pos=0;
+        for(int i=0;i<n;i++){
+            if(nums[i]){
+                if(i+1<n && nums[i]==nums[i+1]){
+                    ans[pos++]=nums[i]*2;
+                    i++;
+                }
+                else{
+                    ans[pos++]=nums[i];
+                }
+            }
+        }
+        return ans;
+    }
+};
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+* stuff you should look before submission
+* constraint and time limit
+* int overflow
+* special test case (n=0||n=1||n=2)
+* don't get stuck on one approach if you get wrong answer
+*/

LeetCode/2461.Maximum_Sum_of_Distinct_Subarrays_With_Length_K.cpp

@@ -0,0 +1,66 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+
+class Solution
+{
+public:
+    long long maximumSubarraySum(vector<int> &nums, int k)
+    {
+        int n=nums.size();
+        long long sum=0,maxo=0;
+
+        map<int,int> check;
+        for(int i=0;i<k-1;i++){
+            sum+=nums[i];
+            check[nums[i]]++;
+        }
+        for(int i=k-1;i<n;i++){
+            check[nums[i]]++;
+            sum+=nums[i];
+            if(check.size()==k){
+                maxo=max(maxo,sum);
+            }
+            if(check[nums[i-k+1]]==1){
+                check.erase(nums[i-k+1]);
+            }
+            else{
+                check[nums[i-k+1]]--;
+            }
+            sum-=nums[i-k+1];
+        }
+        return maxo;
+    }
+};
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+* stuff you should look before submission
+* constraint and time limit
+* int overflow
+* special test case (n=0||n=1||n=2)
+* don't get stuck on one approach if you get wrong answer
+*/

LeetCode/2462.Total_Cost_to_Hire_K_Workers.cpp

@@ -0,0 +1,80 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+class Solution
+{
+public:
+    long long totalCost(vector<int> &costs, int k, int candidates)
+    {
+        int n = costs.size();
+
+        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
+        vector<bool> check(n,false);
+
+        int s=0,e=n-1;
+        for(int i=0;i<candidates;i++)
+        {
+            if(s<=e){
+                pq.push({costs[i],i});
+                s=i+1;
+            }
+        }
+        for(int i=n-1;i>=n-candidates;i--)
+        {
+            if(s<=e){
+                pq.push({costs[i],i});
+                e=i-1;
+            }
+        }
+
+        long long ans=0;
+        for(int i=0;i<k;i++){
+            auto temp1=pq.top();
+            ans+=temp1.first;
+            pq.pop();
+            check[temp1.second]=true;
+            if(temp1.second<s && s<=e){
+                pq.push({costs[s],s});
+                s++;
+            }
+            else if(temp1.second>e && s<=e){
+                pq.push({costs[e],e});
+                e--;
+            }
+        }
+        return ans;
+    }
+};
+
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+ * stuff you should look before submission
+ * constraint and time limit
+ * int overflow
+ * special test case (n=0||n=1||n=2)
+ * don't get stuck on one approach if you get wrong answer
+ */