0(N) Solution
\n# PLS UPVOTE IF YOU LIKE MY SOLUTION\n# Approach\nModify the solve function to take nums1, nums2, and the current index index as parameters. Also, pass a bool
\n# PLS UPVOTE IF YOU LIKE MY SOLUTION\n# Approach\nModify the solve function to take nums1, nums2, and the current index index as parameters. Also, pass a boolean variable swapped to keep track of whether the elements are swapped at the current index.\n\nAdd a base case to the solve function. When the current index reaches the end of the arrays (i.e., index == nums1.size()), return 0 as there are no more elements to consider.\n\nAdd a memoization check in the solve function to avoid redundant calculations. If the result for the current index and swapped state is already calculated, return it from the memoization table dp.\n\nInitialize the variable ans to store the minimum number of swaps required. Set it to INT_MAX initially.\n\nCalculate the previous elements prev1 and prev2 based on the swapped flag and the elements at the previous index.\n\nCheck if swapping is not required (nums1[index] > prev1 and nums2[index] > prev2). In this case, make a recursive call to solve with the next index and the swapped flag set to 0. Update ans with the returned value.\n\nCheck if swapping is required (nums1[index] > prev2 and nums2[index] > prev1). In this case, make a recursive call to solve with the next index and the swapped flag set to 1. Add 1 to the returned value and update ans with the minimum value between ans and the sum.\n\nStore the result in the memoization table dp[index][swapped] and return ans.\n\nIn the minSwap function, insert a dummy element (-1) at the beginning of both nums1 and nums2. This is done to match the indices with the solve function.\n\nCreate a memoization table dp with dimensions [n][2], where n is the size of the modified arrays nums1 and nums2. Initialize all values with -1.\n\nSet the swapped flag to 0 initially.\n\nCall the solve function with nums1, nums2, starting index 1, swapped flag, and the memoization table dp.\n\nReturn the result.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:0(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:0(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\nint solve(vector<int>& nums1, vector<int>& nums2,int index,bool swapped,vector<vector<int>>&dp)\n{\n    if(index==nums1.size())\n    return 0;\n    if(dp[index][swapped]!=-1)\n    {\n        return dp[index][swapped];\n    }\n    int ans=INT_MAX;\n    int prev1=nums1[index-1];\n    int prev2=nums2[index-1];\n    if(swapped)\n    swap(prev1,prev2);\n\n    if(nums1[index]>prev1 && nums2[index]>prev2 )\n    {\n     ans =solve(nums1,nums2,index+1,0,dp);\n    }\n    if(nums1[index]>prev2 && nums2[index]>prev1)\n    {\n        ans =min(ans,1+solve(nums1,nums2,index+1,1,dp));\n    }\n    return dp[index][swapped]=ans;\n\n}\n    int minSwap(vector<int>& nums1, vector<int>& nums2) {\n        nums1.insert(nums1.begin(),-1);\n        nums2.insert(nums2.begin(),-1);\n        int n = nums1.size();\n        vector<vector<int>>dp(n,vector<int>(2,-1));\n        bool swapped=0;\n\n        return solve(nums1,nums2,1,swapped,dp); \n    }\n};\n```\n# PLS HELP ME IN THIS .THIS IS 0(1) SOLUTION BUT IM NOT ABLE TO FIX THE ERROR \n# Code\n```\nclass Solution {\npublic:\nbool solve(vector<int>nums , int n , int t)\n{\n    vector<int>curr(t+1,0);\n    vector<int>next(t+1,0);\n    curr[0]=1;\n    next[0]=1;\n\n    for (int index=n-1;index>=0;index--)\n    {\n        for(int target=0;target<=t;target++)\n        {\n            bool include =false;\n            if(target-nums[index]>=0)\n            {\n                include= next[target-nums[index]];\n            }\n            bool exclude = next[target];\n            curr[target]=include or exclude;\n        }\n        next =curr;\n    }\n    return next[t];\n\n}\n    bool canPartition(vector<int>& nums) {\n        int n = nums.size();\n        int target=0;\n        for(int i =0;i<n;i++)\n        {\n            target+=nums[i];\n        }\n        if(target%2==1)return 0;\n        return solve(nums,n,target/2);\n    }\n};\n```
['Array', 'Dynamic Programming', 'C++']