Detailed Explanation : O(n) Solution using 2 cases for Multisequences DP | Javascript | C++
Intuition\nDynamic Programming Approach:\nThe idea is to keep track of two states at each index i: swap[i] and not_swap[i].\n\nswap[i] represents the minimum nu
# Intuition\n**Dynamic Programming Approach:**\nThe idea is to keep track of two states at each index i: swap[i] and not_swap[i].\n\n`swap[i]` represents the minimum number of swaps required to make `A[0...i]` and `B[0...i]` strictly increasing, considering that you swap elements at index i.\n\n`not_swap[i]` represents the minimum number of swaps required to make `A[0...i]` and `B[0...i]` strictly increasing, considering that you do not swap elements at index i.\n\n\n**Transition Rules:**\n\nIf `A[i-1] < A[i] and B[i-1] < B[i]`, you can keep the same arrangement or swap elements. So, `swap[i] = swap[i-1] + 1 and not_swap[i] = not_swap[i-1]`.\n\nIf `A[i-1] < B[i] and B[i-1] < A[i]`, you can swap at index i or not swap. So, `swap[i] = min(swap[i]`, `not_swap[i-1] + 1)` and n`ot_swap[i] = min(not_swap[i], swap[i-1])`\n\n# Approach\n\nCase 1 : `A[i] > A[i-1] and B[i] > B[i-1]`\n\n`swap[i] = swap[i-1] + 1 `\nif you swapped at i - 1 then swap at i \n\n`not_swap[i] = not_swap[i-1];`\nif you didn\'t swap at i - 1 dont swap here as well\n\n\nCase 2 : `A[i] > B[i-1] and B[i] > A[i-1]`\n\nBoth may not be sorted or one is not sorted in this case\n\nif swapped at [i - 1] index dont swap again otherwise swap here as only one of the 2 swaps are required\n`swap[i] = min( swap[i] , not_swap[i-1] + 1 )`\n\n`not_swap[i] = min( not_swap[i] , swap[i-1]  )`\n\nthere can be both possibilities together or singly so its taking a minimum of the possibilities in the process of calculation\n\n# Complexity\n- Time complexity: $$O(N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:  $$O(N)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n\n```javascript []\nvar minSwap = function(A, B) {\n\n        const N = A.length;\n        const swap = new Array(N).fill(1);\n        const not_swap = new Array(N).fill(0);\n\n        for (let i = 1; i < N; i++) {\n            not_swap[i] = swap[i] = N;\n\n            if (A[i - 1] < A[i] && B[i - 1] < B[i]) {\n                swap[i] = swap[i - 1] + 1;\n                not_swap[i] = not_swap[i - 1];\n            }\n\n            if (A[i - 1] < B[i] && B[i - 1] < A[i]) {\n                swap[i] = Math.min(swap[i], not_swap[i - 1] + 1); \n// swapped at [i] th index\n                not_swap[i] = Math.min(not_swap[i], swap[i - 1]); \n// swapped at [i - 1] th index\n            }\n        }\n        return Math.min(swap[N - 1], not_swap[N - 1]);\n\n    \n};\n```\n```cpp []\nclass Solution {\npublic:\n    int minSwap(vector<int>& A, vector<int>& B) {\n\n        int N = A.size();\n        vector<int> swap(N,1) , not_swap(N,0);\n        for( int i = 1 ; i < N ; i++)\n        {\n            not_swap[i] = swap[i] = N;\n\n            if( A[i-1] < A[i] && B[i-1] < B[i] )\n            {\n                swap[i] = swap[i-1] + 1;\n                not_swap[i] = not_swap[i-1];\n            }\n\n            if( A[i-1] < B[i] && B[i-1] < A[i] )\n            {\n                swap[i] = min( swap[i] , not_swap[i-1] + 1 ); // swapped at [i] th index\n                not_swap[i] = min( not_swap[i] , swap[i-1] ); // swapped at [i - 1] th index\n            }\n        }\n        return min( swap[N-1] , not_swap[N-1] );\n    }\n};\n```
['Array', 'Dynamic Programming', 'C++', 'JavaScript']