Leetcode DP Problems Solution Set

02. Algorithms/09. Dynamic Programming/001. Target Sum/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/001. Target Sum/code.cpp

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+class Solution {
+public:
+ //This question is same as count of number of subsets with given difference
+ int countSubsets(vector<int>& nums, int n, int M)
+ {
+ int t[n + 1][M + 1];
+
+ for(int i = 0; i <= n; i++)
+ {
+ for(int j = 0; j <= M; j++)
+ {
+ if(i == 0)
+ t[i][j] = 0;
+ if(j == 0)
+ t[i][j] = 1;
+ }
+ }
+
+ for(int i = 1; i <= n; i++)
+ {
+ for(int j = 0; j <= M; j++)
+ {
+ if(nums[i - 1] <= j)
+ t[i][j] = t[i - 1][j - nums[i - 1]] + t[i - 1][j];
+ else
+ t[i][j] = t[i - 1][j];
+ }
+ }
+
+ return t[n][M];
+ }
+
+ int findTargetSumWays(vector<int>& nums, int target)
+ {
+ int n = nums.size();
+ int sum = 0;
+ for(int i = 0; i < n; i++)
+ sum += nums[i];
+
+ int M = (sum + target)/2;
+ if(sum < abs(target) || (sum + target) % 2 != 0)
+ return 0;
+ return countSubsets(nums, n, M);
+ }
+};

02. Algorithms/09. Dynamic Programming/002. Climbing Stairs/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/002. Climbing Stairs/code.cpp

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+class Solution {
+public:
+ /*vector<int>dp ;
+
+ Solution(): dp(46,-1){
+ dp[0]=1 ;
+ }
+
+ int climbStairs(int n) {
+
+ if(n<0)
+ return 0;
+
+ if(dp[n]!=-1)
+ return dp[n] ;
+
+ return dp[n] = climbStairs(n-1)+climbStairs(n-2);
+ }*/
+
+ int climbStairs(int n)
+ {
+ if(n==1)
+ return 1 ;
+
+ int dp[n+1];
+
+ dp[0] = 1;
+ dp[1] = 1;
+
+ for(int i=2;i<=n;i++)
+ {
+ dp[i] = dp[i-1]+dp[i-2] ;
+ }
+
+ return dp[n] ;
+ }
+};

02. Algorithms/09. Dynamic Programming/003. Best time to Buy and Sell Stock/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/003. Best time to Buy and Sell Stock/code.cpp

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+class Solution {
+public:
+ int maxProfit(vector<int>& prices) {
+ int cost = INT_MAX, sell=0;
+
+ for(int i:prices)
+ {
+ cost = min(cost,i);
+ sell = max(sell,i-cost) ;
+ }
+
+ return sell ;
+ }
+};

02. Algorithms/09. Dynamic Programming/005. House Robber/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/005. House Robber/code.cpp

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+class Solution {
+public:
+ int rob(vector<int>& nums) {
+ int sz = nums.size();
+ vector<int> dp(sz);
+ for (int i = 0; i < sz; i++) {
+ if (i == 0) {
+ dp[i] = nums[0];
+ } else if (i == 1) {
+ dp[i] = max(nums[1], nums[0]);
+ } else {
+ dp[i] = max(dp[i-1], dp[i-2] + nums[i]);
+ }
+ }
+ return dp[sz-1];
+ }
+};

02. Algorithms/09. Dynamic Programming/006. House Robber II/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/006. House Robber II/code.cpp

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+class Solution {
+public:
+ int rob(vector<int>& nums) {
+ int n = nums.size() ;
+ if(n==1)
+ return nums[0] ;
+ if(n==2)
+ return max(nums[0],nums[1]) ;
+
+ int dp1[n-1] ;
+
+ dp1[0] = nums[0] ;
+ dp1[1] = max(nums[0],nums[1]) ;
+ for(int i=2;i<n-1;i++) {
+ dp1[i] = max(dp1[i-1],nums[i]+dp1[i-2]);
+ }
+
+ int dp2[n-1] ;
+
+ dp2[0] = nums[1] ;
+ dp2[1] = max(nums[2],nums[1]) ;
+ for(int i=2;i<n-1;i++) {
+ dp2[i] = max(dp2[i-1],nums[i+1]+dp2[i-2]);
+ }
+
+ return max(dp1[n-2],dp2[n-2]) ;
+ }
+};

02. Algorithms/09. Dynamic Programming/007. Fibonacci Number/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/007. Fibonacci Number/code.cpp

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+class Solution {
+public:
+ int fib(int n) {
+
+ if(n==0 || n==1)
+ return n;
+ int dp[n+1];
+ dp[0] = 0;
+ dp[1] = 1;
+
+ for(int i=2;i<=n;i++)
+ dp[i] = dp[i-1]+dp[i-2] ;
+
+ return dp[n] ;
+ }
+};

02. Algorithms/09. Dynamic Programming/008. Nth Tribonacci Number/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/008. Nth Tribonacci Number/code.cpp

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+class Solution {
+public:
+ int tribonacci(int n) {
+ if(n==0 || n==1)
+ return n;
+
+ if(n==2)
+ return 1;
+
+ int dp[n+1];
+
+ dp[0]=0;dp[1]=1;dp[2]=1;
+
+ for(int i=3;i<=n;i++)
+ dp[i] = dp[i-1]+dp[i-2]+dp[i-3] ;
+
+ return dp[n];
+ }
+};

02. Algorithms/09. Dynamic Programming/009. Min Cost Climbing Stairs/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/009. Min Cost Climbing Stairs/code.cpp

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+class Solution {
+public:
+
+ int min(int a, int b) {
+ return a<b?a:b ;
+ }
+
+ int solve(vector<int>cost, vector<int>&dp, int index) {
+ if(index>=cost.size())
+ return 0 ;
+
+ if(dp[index]!=-1)
+ return dp[index] ;
+
+ return dp[index] = cost[index] + min(solve(cost,dp,index+1), solve(cost,dp,index+2)) ;
+ } 
+
+ int minCostClimbingStairs(vector<int>& cost) {
+ int n = cost.size() ;
+
+ /*vector<int>dp(n+1,-1) ;
+ return min(solve(cost,dp,0), solve(cost,dp,1)) ;*/
+
+ int dp[n] ;
+
+ dp[0] = cost[0] ;
+ dp[1] = cost[1] ;
+
+ for(int i=2;i<n;i++) {
+ dp[i] = cost[i] + min(dp[i-1],dp[i-2]) ;
+ }
+
+ return min(dp[n-1],dp[n-2]) ;
+ }
+};

02. Algorithms/09. Dynamic Programming/010. Delete and Earn/README.md

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+## Set of Leetcode Problems based on basic Dynamic Programming Approach
+
+# Problem Link - https://leetcode.com/study-plan/dynamic-programming/?progress

02. Algorithms/09. Dynamic Programming/010. Delete and Earn/code.cpp

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+class Solution {
+public:
+ int deleteAndEarn(vector<int>& nums) {
+ vector<int> cnt(10002);
+for(int i=0;i<nums.size();i++)
+{
+cnt[nums[i]]+=nums[i];
+}
+
+vector<int> dp(10002);
+dp[1]=cnt[1];
+dp[2]=max(cnt[1],cnt[2]);
+for(int i=3;i<dp.size();i++)
+{
+dp[i]=max(dp[i-2]+cnt[i],dp[i-1]);
+}
+
+return max(dp[10000],dp[9999]);
+ }
+};