added minimum-number-of-k-consecutive-bit-flips.java code

Author: pranav-3005

Coding/Java/minimum-number-of-k-consecutive-bit-flips.java

@@ -0,0 +1,52 @@
+//Question : https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/description/
+
+//Question exp : convert alll 0 -> 1 by doing k bit flip
+
+//Ans approach : greedy approach
+class Solution {
+    public int minKBitFlips(int[] nums, int k) {
+        int n=nums.length;
+        int flipCount=0; // range [cur - (k-1)elem] will be flipped this much times
+        int totalFlips=0 ; //final ans
+
+        //true will be marked for all range's(cur indx - k-1th indx) end elem.
+        //While traversing, if visited[cur elem]=true,
+        //then we reduce the flipCount of cur_range ,as we move towards nxt range
+        boolean[] visited=new boolean[n];
+
+        for(int i=0;i<n;i++)
+        {
+            //only works if, (which is flipped flipCount no of times) cur_elem's value = 0
+                // 0 flipped even times = 0
+                //1 flipped odd times = 0
+            if( (nums[i]==0 && flipCount%2==0) || (nums[i]==1 && flipCount%2!=0) )
+            {
+                flipCount++;
+                totalFlips++;
+
+                if( i+(k-1) < n )
+                {
+                    //mark end of cur range as true
+                    //so, we can rmv the flipCount of this range, after reaching this endPoint.
+                    visited[i+(k-1)]=true;
+                }
+                else
+                {
+                    //if end elem goes out of range
+                    //hence, all of arr elems cannot be flipped to 1
+                    return -1; 
+                }
+            }
+
+            if(visited[i]==true)
+            {
+                //end point of cur_range is reached.
+                //so reduce the flipCount of cur_range 
+                flipCount--;
+            }
+
+        } 
+
+        return totalFlips;
+    }
+}