From edaa0852ca5a9d159afb21c10321b6da840179c5 Mon Sep 17 00:00:00 2001
From: anebz <anebzt@protonmail.com>
Date: Fri, 20 Mar 2020 16:56:04 +0100
Subject: [PATCH] 5.3. flip bit to win v1

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+# 5.3. Flip Bit to Win
+
+Hints: #358,#375,#390
+
+> You have an integer and you can flip exactly one bit from a 0 to a 1. Write code to find the length of the longest sequence of 1s you could create. Input: 1775, or  11011101111, Output: 8
+
+Brute force idea: find all 0s, flip all of them, count all 1s sequences (how?), output max.
+
+Another idea: find longest sequence of 1s as of now, flip adjacent 0 and count length, flip other 0 (if not end/beginning of sequence) and count length, output max.
+
+But how to find first 0? and count consecutive 1s? And how to make it efficient?
+
+## Hint 1
+
+> Start with a brute force solution for each
+
+Flipping function, need the position of the 0 bit.
+
+```c++
+boolean setBit(int num, int i) {
+    return num | (1 << i);
+}
+```
+
+[Length of the Longest Consecutive 1s in Binary Representation](https://www.geeksforgeeks.org/length-longest-consecutive-1s-binary-representation/). Bit magic.
+
+```c++
+#include<bits/stdc++.h>
+using namespace std;
+  
+int maxConsecutiveOnes(int num) {
+    // Initialize result
+    int count = 0;
+  
+    // Count the number of iterations to reach num = 0.
+    while (num!=0) {
+        // This operation reduces length
+        // of every sequence of 1s by one.
+        num = (num & (num << 1));
+        count++;
+    }
+  
+    return count;
+}
+```
+
+```c++
+  11101111   (x)
+& 11011110   (x << 1)
+----------
+  11001110   (x & (x << 1))
+    ^    ^
+    |    |
+trailing 1 removed
+```
+
+The operation x = (x & (x << 1)) reduces length of every sequence of 1s by one in binary representation of x. If we keep doing this operation in a loop, we end up with x = 0. The number of iterations required to reach 0 is the length of the longest consecutive sequence of 1s.
+
+If I find all 0s, flip them, I can use this to find the longest sequence.
+
+[Number of leading zeros in binary representation of a given number](https://www.geeksforgeeks.org/number-of-leading-zeros-in-binary-representation-of-a-given-number/).
+
+```c++
+#include <vector>
+using namespace std;
+
+bool findPosof0(int num) {
+  int i = sizeof(num) * 8; // number of bits in num
+  vector<int> positions;
+
+  while (i--) {
+    // checking if i th bit of num is 1
+    if (!(num & (1 << i))) {
+      positions.push_back(i);
+    }
+    num = (num << 1);
+  }
+  // found first 0
+  return positions;
+}
+
+int num = 17;
+int max1s = 0;
+// finds all positions of 0
+vector<int> positions0 = findPosOf0(num);
+
+for (pos : positions0) {
+  // flip the first 0, returns a boolean so don't know if this works
+  num = setBit(num, first0);
+  max1s = max(max1s, maxConsecutiveOnes(num));
+}
+
+// count consecutive 1s
+cout << max1s << endl;
+```
+
+Time complexity O(2n) and space complexity O(2n).
+
+## Hint 2
+
+> Get Next: Picture a binary number-something with a bunch of 1s and Os spread out throughout the number. Suppose you flip a 1 to a 0 and a 0 to a 1. In what case will the number get bigger? In what case will it get smaller?
+
+Flipping 1 to 0 makes it smaller, and vice versa.
+
+## Hint 3
+
+> If you flip a 1 to a 0 and a 0 to a 1, it will get bigger if the 0->1 bit is more significant than the 1->0 bit. How can you use this to create the next biggest number (with the same number of 1s)?
+
+I don't know.
+
+## Hint 4
+
+> When you do a binary subtraction, you flip the rightmost 0s to a 1, stopping when you get to a 1 (which is also flipped). Everything (all the 1 s and 0s) on the left will stay put.
+
+Yes. And in reverse, if we have a bunch of 11111s, if we add one, it will create many 0s.
+
+## Hint 5
+
+> Flipping a 0 to a 1 will create a bigger number. The farther right the index is the smaller the bigger number is. If we have a number like 1001, we want to flip the rightmost 0 (to create 1011). But if we have a number like 1010, we should not flip the rightmost 1.
+
+Sure but we're not looking for the biggest , rather for the longest sequence of 1s. If we have 10111011, we should flip the second 0. In this case, flipping any 0 creates a sequence of two 1s. So it shouldn't matter :thinking:.
+
+## Hint 6
+
+> 
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