Runtime: 1198 ms (Top 36.1%) | Memory: 149.20 MB (Top 27.6%)

Author: AnasImloul

scripts/algorithms/M/Minimum Swaps To Make Sequences Increasing/Minimum Swaps To Make Sequences Increasing.py

@@ -1,48 +1,24 @@
+# Runtime: 1198 ms (Top 36.1%) | Memory: 149.20 MB (Top 27.6%)
+
 class Solution:
     def minSwap(self, nums1: List[int], nums2: List[int]) -> int:
-        """
+        dp = [[-1]*2 for i in range(len(nums1))]
 
-        
-        q. is there a greedy wat to know tht shall i switch or not?
-        a. 
-            [1, 3, 6] [1, 4, 4]
+        def solve(prev1, prev2, i, swaped):
+            if i >= len(nums1): return 0
+            if dp[i][swaped] != -1: return dp[i][swaped]
+                        
+            ans = 2**31
 
-            at i=1, we cannot greedily determine about switching.. we need lookahead, i.e DP
-			
-			at each (i) we have a choice
-            if its currently not increasing:
-                we have to see if switching will make it increasing
-            else:
-                we can move to i+1 since its already incr
-                OR
-                if switch is possible, we switch and cont.
-        
-        possible complexity:
-            dp(i, prev1, prev2) :
-        """
-        
-        @cache
-        def dp(i, preva, prevb):
+            # No Swap
+            if nums1[i] > prev1 and nums2[i] > prev2:
+                ans = solve(nums1[i], nums2[i], i+1, 0) 
 
-            if i == len(nums1):
-                return 0
+            # Swap
+            if nums1[i] > prev2 and nums2[i] > prev1:
+                ans = min(ans, 1 + solve(nums2[i], nums1[i], i+1, 1)) 
 
-            if nums1[i] > preva and nums2[i] > prevb: 
-                # we are increasing as is, we have 2 choices now, make swap if possible else dont
-                if nums1[i] > prevb and nums2[i] > preva: 
-                    # switch is also possible
-                    return min(
-                        1 + dp(i+1, nums2[i], nums1[i]),
-                        dp(i+1, nums1[i], nums2[i])
-                    )
-                else:
-                    return dp(i+1, nums1[i], nums2[i])
-            else:
-                # we only have 1 choice
-                if nums1[i] > prevb and nums2[i] > preva:
-                    # switch is possible
-                    return 1 + dp(i+1, nums2[i], nums1[i]) # swithced
-                else:
-                    return inf
-            
-        return min(dp(1, nums1[0], nums2[0]), 1+dp(1, nums2[0], nums1[0]))
+            dp[i][swaped] = ans
+            return ans
+        
+        return solve(-1, -1, 0, 0)