added cses

leetcode/lc1888/SlidingWindow.cpp

@@ -0,0 +1,56 @@
+class Solution {
+public:
+    int minFlips(string s) {
+        int n = s.size();
+        string t = s + s;
+
+        int ans = INT_MAX;
+        int diff1=0, diff2 = 0;
+        for(int i =0;i<t.size();i++){
+            char expected1 = (i%2 ? '1':'0');
+            char expected2 = (i%2 ? '0':'1');
+
+            if(t[i]!=expected1) diff1++;
+            if(t[i]!=expected2) diff2++;
+
+            if(i>=n){
+                char old = t[i-n];
+                char oldExpected1 = ((i-n)%2 ? '1' : '0');
+                char oldExpected2 = ((i-n)%2 ? '0' : '1');
+
+                if(old!=oldExpected1) diff1--;
+                if(old!=oldExpected2) diff2--;
+
+             
+            }
+               if(i>=n-1)
+                    ans  = min(ans, min(diff1,diff2));
+        
+        }
+        return ans;
+    }
+};
+
+//SC : o(n)  TC : o(n)
+
+
+/*
+Approach:
+Since the string can be rotated, we simulate all possible rotations by
+doubling the string (t = s + s). Any rotation of s will appear as a
+substring of length n in this doubled string. We then slide a window of
+size n across t and compute how many flips are required to make the
+current window alternating.
+
+There are only two possible alternating patterns: "010101..." and
+"101010...". We maintain two mismatch counters (diff1 and diff2) that
+track how many characters in the current window differ from each pattern.
+As the window expands, we add the mismatch contribution of the new
+character. When the window exceeds size n, we remove the contribution of
+the character that leaves the window.
+
+For every window of size n, the minimum of diff1 and diff2 gives the
+number of flips required for that rotation. We keep track of the global
+minimum across all windows.
+
+*/