Find zeroes to be flipped so that number of consecutive 1’s is maximized
Given a binary array and an integer m, find the position of zeroes flipping which creates maximum number of consecutive 1’s in array.
Examples :
Input: arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1} m = 2 Output: 5 7 We are allowed to flip maximum 2 zeroes. If we flip arr[5] and arr[7], we get 8 consecutive 1's which is maximum possible under given constraints Input: arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1} m = 1 Output: 7 We are allowed to flip maximum 1 zero. If we flip arr[7], we get 5 consecutive 1's which is maximum possible under given constraints. Input: arr[] = {0, 0, 0, 1} m = 4 Output: 0 1 2 Since m is more than number of zeroes, we can flip all zeroes.
Source: http://www.careercup.com/question?id=5106425965576192
A Simple Solution is to consider every subarray by running two loops. For every subarray, count number of zeroes in it. Return the maximum size subarray with m or less zeroes. Time Complexity of this solution is O(n2).
A Better Solution is to use auxiliary space to solve the problem in O(n) time.
For all positions of 0’s calculate left[] and right[] which defines the number of consecutive 1’s to the left of i and right of i respectively.
For example, for arr[] = {1, 1, 0, 1, 1, 0, 0, 1, 1, 1} and m = 1, left[2] = 2 and right[2] = 2, left[5] = 2 and right[5] = 0, left[6] = 0 and right[6] = 3.
left[] and right[] can be filled in O(n) time by traversing array once and keeping track of last seen 1 and last seen 0. While filling left[] and right[], we also store indexes of all zeroes in a third array say zeroes[]. For above example, this third array stores {2, 5, 6}
Now traverse zeroes[] and for all consecutive m entries in this array, compute the sum of 1s that can be produced. This step can be done in O(n) using left[] and right[].
C++
// C++ program to find positions of zeroes flipping which // produces maximum number of consecutive 1's #include <bits/stdc++.h> using namespace std; // m is maximum of number zeroes allowed to flip // n is size of array vector< int > maximized_one( int arr[], int n, int m) { // Left array int left[n] = { 0 }; // Right array int right[n] = { 0 }; // Array will contain zeroes position vector< int > zero_pos; // Stores count int count = 0; int previous_index_of_zero = -1; for ( int i = 0; i < n; i++) { if (arr[i]) { count++; } else { left[i] = count; zero_pos.push_back(i); if (previous_index_of_zero != i && previous_index_of_zero != -1) { right[previous_index_of_zero] = count; } count = 0; // To keep track of the previous index of zeroes previous_index_of_zero = i; } } right[previous_index_of_zero] = count; int max_one = -1; vector< int > result_index; int i = 0; while (i <= (zero_pos.size()) - m) { int temp = 0; vector< int > index; for ( int c = 0; c < m; c++) { temp += left[zero_pos[i + c]] + right[zero_pos[i + c]] + 1; // Index is updated index.push_back(zero_pos[i + c]); } // Decrement temp by m-1 because when we are // calculating temp we are adding 1 in it. So, in // order to get exact count of 1. This decrement is // applicable only when value of m is greater than 1 temp = temp - (m - 1); // Updating max value when we get the new max temp // and result_index as well if (temp > max_one) { max_one = temp; result_index = index; } i += 1; } return result_index; } // Driver program int main() { int arr[] = { 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1 }; int m = 2; int n = sizeof (arr) / sizeof (arr[0]); cout << "Index of zeroes that are flipped: " ; vector< int > result_index = maximized_one(arr, n, m); for ( auto i : result_index) { cout << i << " " ; } return 0; } |
Java
// Java program to find positions of zeroes flipping which // produces maximum number of consecutive 1's import java.util.*; class GFG{ // m is maximum of number zeroes allowed to flip // n is size of array static Vector<Integer> maximized_one( int arr[], int n, int m) { // Left array int left[] = new int [n]; // Right array int right[] = new int [n]; // Array will contain zeroes position Vector<Integer> zero_pos = new Vector<>(); // Stores count int count = 0 ; int previous_index_of_zero = - 1 ; for ( int i = 0 ; i < n; i++) { if (arr[i]!= 0 ) { count++; } else { left[i] = count; zero_pos.add(i); if (previous_index_of_zero != i && previous_index_of_zero != - 1 ) { right[previous_index_of_zero] = count; } count = 0 ; // To keep track of the previous index of zeroes previous_index_of_zero = i; } } right[previous_index_of_zero] = count; int max_one = - 1 ; Vector<Integer> result_index = new Vector<>(); int i = 0 ; while (i <= (zero_pos.size()) - m) { int temp = 0 ; Vector<Integer> index = new Vector<>(); for ( int c = 0 ; c < m; c++) { temp += left[zero_pos.elementAt(i + c)] + right[zero_pos.elementAt(i + c)] + 1 ; // Index is updated index.add(zero_pos.elementAt(i + c)); } // Decrement temp by m-1 because when we are // calculating temp we are adding 1 in it. So, in // order to get exact count of 1. This decrement is // applicable only when value of m is greater than 1 temp = temp - (m - 1 ); // Updating max value when we get the new max temp // and result_index as well if (temp > max_one) { max_one = temp; result_index = index; } i += 1 ; } return result_index; } // Driver program public static void main(String[] args) { int arr[] = { 1 , 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 1 , 1 }; int m = 2 ; int n = arr.length; System.out.print( "Index of zeroes that are flipped: [" ); Vector<Integer> result_index = maximized_one(arr, n, m); for ( int i : result_index) { System.out.print(i+ " " ); } System.out.print( "]" ); } } // This code contributed by umadevi9616 |
Python3
# Python3 code for the above approach def maximized_one(arr,n,m): # Left array left = [ 0 ] * n # Right array right = [ 0 ] * n # Array will contain zeroes position zero_pos = [] # Stores count count = 0 previous_index_of_zero = - 1 for i in range (n): if arr[i] = = 1 : count + = 1 if arr[i] = = 0 : left[i] = count zero_pos.append(i) if previous_index_of_zero ! = i and previous_index_of_zero! = - 1 : right[previous_index_of_zero] = count count = 0 # To keep track of the previous index of zeroes previous_index_of_zero = i right[previous_index_of_zero] = count # print(left) # print(right) # print(zero_pos) max_one = - 1 result_index = 0 i = 0 while (i< = len (zero_pos) - m): temp = 0 index = [] for c in range (m): # print(zero_pos[i+c],left[zero_pos[i+c]],right[zero_pos[i+c]]) temp + = left[zero_pos[i + c]] + right[zero_pos[i + c]] + 1 # Index is updated index.append(zero_pos[i + c]) # Decrement temp by m-1 because when we are calculating temp # we are adding 1 in it. # So, in order to get exact count of 1. # This decrement is applicable only when value of m is greater than 1 temp = temp - (m - 1 ) # Updating max value when we get the new max temp # and result_index as well if temp > max_one: max_one = temp result_index = index i + = 1 return result_index # Driver Code if __name__ = = '__main__' : arr = [ 1 , 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 1 , 1 ] n = len (arr) m = 2 print ( 'Index of zeroes that are flipped: ' ,maximized_one(arr,n,m)) |
C#
// C# program to find positions of zeroes flipping which // produces maximum number of consecutive 1's using System; using System.Collections.Generic; public class GFG { // m is maximum of number zeroes allowed to flip // n is size of array static List< int > maximized_one( int []arr, int n, int m) { // Left array int []left = new int [n]; // Right array int []right = new int [n]; // Array will contain zeroes position List< int > zero_pos = new List< int >(); // Stores count int count = 0; int previous_index_of_zero = -1; for ( int j = 0; j < n; j++) { if (arr[j] != 0) { count++; } else { left[j] = count; zero_pos.Add(j); if (previous_index_of_zero != j && previous_index_of_zero != -1) { right[previous_index_of_zero] = count; } count = 0; // To keep track of the previous index of zeroes previous_index_of_zero = j; } } right[previous_index_of_zero] = count; int max_one = -1; List< int > result_index = new List< int >(); int i = 0; while (i <= (zero_pos.Count) - m) { int temp = 0; List< int > index = new List< int >(); for ( int c = 0; c < m; c++) { temp += left[zero_pos[i + c]] + right[zero_pos[i + c]] + 1; // Index is updated index.Add(zero_pos[i + c]); } // Decrement temp by m-1 because when we are // calculating temp we are adding 1 in it. So, in // order to get exact count of 1. This decrement is // applicable only when value of m is greater than 1 temp = temp - (m - 1); // Updating max value when we get the new max temp // and result_index as well if (temp > max_one) { max_one = temp; result_index = index; } i += 1; } return result_index; } // Driver program public static void Main(String[] args) { int []arr = { 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1 }; int m = 2; int n = arr.Length; Console.Write( "Index of zeroes that are flipped: [" ); List< int > result_index = maximized_one(arr, n, m); foreach ( int i in result_index) { Console.Write(i+ " " ); } Console.Write( "]" ); } } // This code is contributed by umadevi9616 |
Javascript
<script> // javascript program to find positions of zeroes flipping which // produces maximum number of consecutive 1's // m is maximum of number zeroes allowed to flip // n is size of array function maximized_one(arr, n, m) { // Left array var left = Array(n).fill(0); // Right array var right = Array(n).fill(0); // Array will contain zeroes position var zero_pos = new Array(); // Stores count var count = 0; var previous_index_of_zero = -1; for ( var i = 0; i < n; i++) { if (arr[i] != 0) { count++; } else { left[i] = count; zero_pos.push(i); if (previous_index_of_zero != i && previous_index_of_zero != -1) { right[previous_index_of_zero] = count; } count = 0; // To keep track of the previous index of zeroes previous_index_of_zero = i; } } right[previous_index_of_zero] = count; var max_one = -1; var result_index = Array(); var i = 0; while (i <= (zero_pos.length) - m) { var temp = 0; var index = Array(); for ( var c = 0; c < m; c++) { temp += left[zero_pos[i + c]] + right[zero_pos[i + c]] + 1; // Index is updated index.push(zero_pos[i + c]); } // Decrement temp by m-1 because when we are // calculating temp we are adding 1 in it. So, in // order to get exact count of 1. This decrement is // applicable only when value of m is greater than 1 temp = temp - (m - 1); // Updating max value when we get the new max temp // and result_index as well if (temp > max_one) { max_one = temp; result_index = index; } i += 1; } return result_index; } // Driver program var arr = [ 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1 ]; var m = 2; var n = arr.length; document.write( "Index of zeroes that are flipped: [" ); var result_index = maximized_one(arr, n, m); for ( var i = 0; i < result_index.length; i++) { document.write(result_index[i] + " " ); } document.write( "]" ); // This code is contributed by gauravrajput1 </script> |
Index of zeroes that are flipped: [5, 7]
An Efficient Solution can solve the problem in O(n) time and O(1) space. The idea is to use Sliding Window for the given array. The solution is taken from here.
Let us use a window covering from index wL to index wR. Let the number of zeros inside the window be zeroCount. We maintain the window with at most m zeros inside.
The main steps are:
– While zeroCount is no more than m: expand the window to the right (wR++) and update the count zeroCount.
– While zeroCount exceeds m, shrink the window from left (wL++), update zeroCount;
– Update the widest window along the way. The positions of output zeros are inside the best window.
Below is the implementation of the idea.
C++
// C++ program to find positions of zeroes flipping which // produces maximum number of consecutive 1's #include<bits/stdc++.h> using namespace std; // m is maximum of number zeroes allowed to flip // n is size of array void findZeroes( int arr[], int n, int m) { // Left and right indexes of current window int wL = 0, wR = 0; // Left index and size of the widest window int bestL = 0, bestWindow = 0; // Count of zeroes in current window int zeroCount = 0; // While right boundary of current window doesn't cross // right end while (wR < n) { // If zero count of current window is less than m, // widen the window toward right if (zeroCount <= m) { if (arr[wR] == 0) zeroCount++; wR++; } // If zero count of current window is more than m, // reduce the window from left if (zeroCount > m) { if (arr[wL] == 0) zeroCount--; wL++; } // Update widest window if this window size is more if ((wR-wL > bestWindow) && (zeroCount<=m)) { bestWindow = wR-wL; bestL = wL; } } // Print positions of zeroes in the widest window for ( int i=0; i<bestWindow; i++) { if (arr[bestL+i] == 0) cout << bestL+i << " " ; } } // Driver program int main() { int arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1}; int m = 2; int n = sizeof (arr)/ sizeof (arr[0]); cout << "Indexes of zeroes to be flipped are " ; findZeroes(arr, n, m); return 0; } |
Java
//Java to find positions of zeroes flipping which // produces maximum number of consecutive 1's class Test { static int arr[] = new int []{ 1 , 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 1 }; // m is maximum of number zeroes allowed to flip static void findZeroes( int m) { // Left and right indexes of current window int wL = 0 , wR = 0 ; // Left index and size of the widest window int bestL = 0 , bestWindow = 0 ; // Count of zeroes in current window int zeroCount = 0 ; // While right boundary of current window doesn't cross // right end while (wR < arr.length) { // If zero count of current window is less than m, // widen the window toward right if (zeroCount <= m) { if (arr[wR] == 0 ) zeroCount++; wR++; } // If zero count of current window is more than m, // reduce the window from left if (zeroCount > m) { if (arr[wL] == 0 ) zeroCount--; wL++; } // Update widest window if this window size is more if ((wR-wL > bestWindow) && (zeroCount<=m)) { bestWindow = wR-wL; bestL = wL; } } // Print positions of zeroes in the widest window for ( int i= 0 ; i<bestWindow; i++) { if (arr[bestL+i] == 0 ) System.out.print(bestL+i + " " ); } } // Driver method to test the above function public static void main(String[] args) { int m = 2 ; System.out.println( "Indexes of zeroes to be flipped are " ); findZeroes(m); } } |
Python3
# Python3 program to find positions # of zeroes flipping which produces # maximum number of consecutive 1's # m is maximum of number zeroes allowed # to flip, n is size of array def findZeroes(arr, n, m) : # Left and right indexes of current window wL = wR = 0 # Left index and size of the widest window bestL = bestWindow = 0 # Count of zeroes in current window zeroCount = 0 # While right boundary of current # window doesn't cross right end while wR < n: # If zero count of current window is less than m, # widen the window toward right if zeroCount < = m : if arr[wR] = = 0 : zeroCount + = 1 wR + = 1 # If zero count of current window is more than m, # reduce the window from left if zeroCount > m : if arr[wL] = = 0 : zeroCount - = 1 wL + = 1 # Update widest window if # this window size is more if (wR - wL > bestWindow) and (zeroCount< = m) : bestWindow = wR - wL bestL = wL # Print positions of zeroes # in the widest window for i in range ( 0 , bestWindow): if arr[bestL + i] = = 0 : print (bestL + i, end = " " ) # Driver program arr = [ 1 , 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 1 ] m = 2 n = len (arr) print ( "Indexes of zeroes to be flipped are" , end = " " ) findZeroes(arr, n, m) # This code is contributed by Shreyanshi Arun. |
C#
// C# to find positions of zeroes flipping which // produces maximum number of consecutive 1's using System; class Test { static int []arr = new int []{1, 0, 0, 1, 1, 0, 1, 0, 1, 1}; // m is maximum of number zeroes allowed to flip static void findZeroes( int m) { // Left and right indexes of current window int wL = 0, wR = 0; // Left index and size of the widest window int bestL = 0, bestWindow = 0; // Count of zeroes in current window int zeroCount = 0; // While right boundary of current // window doesn't cross right end while (wR < arr.Length) { // If zero count of current window is less // than m, widen the window toward right if (zeroCount <= m) { if (arr[wR] == 0) zeroCount++; wR++; } // If zero count of current window is more than m, // reduce the window from left if (zeroCount > m) { if (arr[wL] == 0) zeroCount--; wL++; } // Update widest window if this window size is more if ((wR-wL > bestWindow) && (zeroCount<=m)) { bestWindow = wR-wL; bestL = wL; } } // Print positions of zeroes in the widest window for ( int i = 0; i < bestWindow; i++) { if (arr[bestL + i] == 0) Console.Write(bestL + i + " " ); } } // Driver method to test the above function public static void Main(String[] args) { int m = 2; Console.Write( "Indexes of zeroes to be flipped are " ); findZeroes(m); } } // This code is contributed by parashar |
PHP
<?php // PHP program to find positions of // zeroes flipping which produces // maximum number of consecutive 1's // m is maximum of number zeroes // allowed to flip n is size of array function findZeroes( $arr , $n , $m ) { // Left and right indexes // of current window $wL = 0; $wR = 0; // Left index and size of // the widest window $bestL = 0; $bestWindow = 0; // Count of zeroes in // current window $zeroCount = 0; // While right boundary of // current window doesn't cross // right end while ( $wR < $n ) { // If zero count of current // window is less than m, // widen the window toward right if ( $zeroCount <= $m ) { if ( $arr [ $wR ] == 0) $zeroCount ++; $wR ++; } // If zero count of current // window is more than m, // reduce the window from left if ( $zeroCount > $m ) { if ( $arr [ $wL ] == 0) $zeroCount --; $wL ++; } // Update widest window if // this window size is more if (( $wR - $wL > $bestWindow ) && ( $zeroCount <= $m )) { $bestWindow = $wR - $wL ; $bestL = $wL ; } } // Print positions of zeroes // in the widest window for ( $i = 0; $i < $bestWindow ; $i ++) { if ( $arr [ $bestL + $i ] == 0) echo $bestL + $i . " " ; } } // Driver Code $arr = array (1, 0, 0, 1, 1, 0, 1, 0, 1, 1); $m = 2; $n = sizeof( $arr )/sizeof( $arr [0]); echo "Indexes of zeroes to be flipped are " ; findZeroes( $arr , $n , $m ); return 0; // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript program to find positions of // zeroes flipping which produces // maximum number of consecutive 1's // m is maximum of number zeroes // allowed to flip n is size of array function findZeroes(arr, n, m) { // Left and right indexes // of current window let wL = 0; let wR = 0; // Left index and size of // the widest window let bestL = 0; let bestWindow = 0; // Count of zeroes in // current window let zeroCount = 0; // While right boundary of // current window doesn't cross // right end while (wR < n) { // If zero count of current // window is less than m, // widen the window toward right if (zeroCount <= m) { if (arr[wR] == 0) zeroCount++; wR++; } // If zero count of current // window is more than m, // reduce the window from left if (zeroCount > m) { if (arr[wL] == 0) zeroCount--; wL++; } // Update widest window if // this window size is more if ((wR - wL > bestWindow) && (zeroCount <= m)) { bestWindow = wR - wL; bestL = wL; } } // Print positions of zeroes // in the widest window for (let i = 0; i < bestWindow; i++) { if (arr[bestL + i] == 0) document.write(bestL + i + " " ); } } // Driver Code let arr = new Array(1, 0, 0, 1, 1, 0, 1, 0, 1, 1); let m = 2; let n = arr.length; document.write( "Indexes of zeroes to be flipped are " ); findZeroes(arr, n, m); // This code is contributed by Saurabh Jaiswal </script> |
Indexes of zeroes to be flipped are 5 7
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