使左侧全为 1，右侧全为 0 的最小翻转 |设置 2
给定一个二进制数组，我们可以将左侧的所有 1 和右侧的所有 0 翻转。计算使左侧所有 1 和右侧所有 0 所需的最小翻转。
例子 ：
Input: 1011000
Output: 1
1 flip is required to make it 1111000.
Input : 00001
Output : 2
2 flips required to make it 10000.我们在下面的帖子中讨论了基于位掩码的解决方案。使左侧全为 1，右侧全为 0 的最小翻转 |设置 1（使用位掩码）
它可以通过 O(N) 时间复杂度（其中 N – 位数）和 O(N) 额外内存来完成
- 计算从左到右移动时需要完成的“0”翻转次数以使所有“1”位。
- 计算在从右向左移动以使所有位为“0”时需要完成的“1”翻转次数。
- 遍历位之间的所有位置并从两个数组中找到'0'-翻转+'1'-翻转的最小和。
C++
// CPP program to find minimum flips required
// to make all 1s in left and 0s in right.
#include
using namespace std;
int minimalFilps(string bits)
{
int n = bits.length();
// two arrays will keep the count for number
// of 0s' and 1s' to be flipped while
// traversing from left to right and right to
// left respectively
int flipsFromLeft[n];
int flipsFromRight[n];
// Fill flipsFromLeft[]
int flips = 0;
for (int i = 0; i < n; i++) {
if (bits[i] == '0')
flips++;
flipsFromLeft[i] = flips;
}
// Fill flipsFromRight[]
flips = 0;
for (int i = n - 1; i >= 0; i--) {
if (bits[i] == '1')
flips++;
flipsFromRight[i] = flips;
}
// initialize minFlip to highest int value. If sum
// of leftflip and rightFlip is smaller than minflips,
// then update minFlips
int minFlips = INT_MAX;
for (int i = 1; i < n; i++) {
if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
}
return minFlips;
}
// Driver code
int main()
{
string bits = "00001";
cout << minimalFilps(bits) << endl;
return 0;
} Java
// Java program to find minimum flips required
// to make all 1s in left and 0s in right.
import java.io.*;
class GFG
{
static int minimalFilps(String bits)
{
int n = bits.length();
// two arrays will keep the count
// for number of 0s' and 1s' to be
// flipped while traversing from
// left to right and right to
// left respectively
int flipsFromLeft[] = new int[n];
int flipsFromRight[] =new int[n] ;
// Fill flipsFromLeft[]
int flips = 0;
for (int i = 0; i < n; i++)
{
if (bits.charAt(i) == '0')
flips++;
flipsFromLeft[i] = flips;
}
// Fill flipsFromRight[]
flips = 0;
for (int i = n - 1; i >= 0; i--)
{
if (bits.charAt(i) == '1')
flips++;
flipsFromRight[i] = flips;
}
// initialize minFlip to highest int value. If sum
// of leftflip and rightFlip is smaller than minflips,
// then update minFlips
int minFlips = Integer.MAX_VALUE;
for (int i = 1; i < n; i++)
{
if (flipsFromLeft[i - 1] + flipsFromRight[i]
< minFlips)
minFlips = flipsFromLeft[i - 1]
+ flipsFromRight[i];
}
return minFlips;
}
// Driver code
public static void main (String[] args)
{
String bits = "00001";
System.out.println(minimalFilps(bits));
}
}
// This code is contributed by vt_m.Python3
# Python 3 program to find minimum flips required
# to make all 1s in left and 0s in right.
import sys
def minimalFilps(bits):
n = len(bits)
# two arrays will keep the count for number
# of 0s' and 1s' to be flipped while
# traversing from left to right and right to
# left respectively
flipsFromLeft = [0 for i in range(n)]
flipsFromRight = [0 for i in range(n)]
# Fill flipsFromLeft[]
flips = 0
for i in range(0, n, 1):
if (bits[i] == '0'):
flips = flips + 1
flipsFromLeft[i] = flips
# Fill flipsFromRight[]
flips = 0
i = n - 1
while(i >= 0):
if (bits[i] == '1'):
flips = flips + 1
i = i - 1
flipsFromRight[i] = flips
# initialize minFlip to highest int value.
# If sum of leftflip and rightFlip is smaller
# than minflips, then update minFlips
minFlips = sys.maxsize
for i in range(1, n, 1):
if (flipsFromLeft[i - 1] +
flipsFromRight[i] < minFlips):
minFlips = (flipsFromLeft[i - 1] +
flipsFromRight[i])
return minFlips
# Driver code
if __name__ == '__main__':
bits = "00001"
print(minimalFilps(bits))
# This code is contributed by
# Surendra_GangwarC#
// C# program to find minimum flips required
// to make all 1s in left and 0s in right.
using System;
class GFG
{
static int minimalFilps(String bits)
{
int n = bits.Length;
// two arrays will keep the count
// for number of 0s' and 1s' to be
// flipped while traversing from
// left to right and right to
// left respectively
int []flipsFromLeft = new int[n];
int []flipsFromRight =new int[n] ;
// Fill flipsFromLeft[]
int flips = 0;
for (int i = 0; i < n; i++)
{
if (bits[i] == '0')
flips++;
flipsFromLeft[i] = flips;
}
// Fill flipsFromRight[]
flips = 0;
for (int i = n - 1; i >= 0; i--)
{
if (bits[i] == '1')
flips++;
flipsFromRight[i] = flips;
}
// initialize minFlip to highest int value.
// If sum of leftflip and rightFlip is smaller
// than minflips, then update minFlips
int minFlips = int.MaxValue;
for (int i = 1; i < n; i++)
{
if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
}
return minFlips;
}
// Driver code
public static void Main ()
{
string bits = "00001";
Console.WriteLine(minimalFilps(bits));
}
}
// This code is contributed by vt_m.PHP
= 0; $i--)
{
if ($bits[$i] == '1')
$flips++;
$flipsFromRight[$i] = $flips;
}
// initialize minFlip to
// highest int value. If sum
// of leftflip and rightFlip
// is smaller than minflips,
// then update minFlips
$INT_MAX=2147483647;
$minFlips = $INT_MAX;
for ($i = 1; $i < $n; $i++)
{
if ($flipsFromLeft[$i - 1] +
$flipsFromRight[$i] < $minFlips)
$minFlips = $flipsFromLeft[$i - 1] +
$flipsFromRight[$i];
}
return $minFlips;
}
// Driver Code
$bits = "00001";
echo minimalFilps($bits) ;
// This code is contributed by nitin mittal.
?>Javascript
输出：
2