检查所有字符的频率是否可以通过一次删除变得相同

给定一个包含低位字母字符的字符串，我们最多需要从该字符串中删除一个字符，以使每个不同字符在字符串中的频率变得相同。

例子：

Input: str = “xyyz” 
Output: Yes 
We can remove character ’y’ from above 
string to make the frequency of each 
character same. 

Input: str = “xyyzz” 
Output: Yes 
We can remove character ‘x’ from above 
string to make the frequency of each 
character same.

Input: str = “xxxxyyzz” 
Output: No 
It is not possible to make frequency of 
each character same just by removing at 
most one character from above string.

方法：这个问题可以使用散列的概念来解决。在这个问题中要注意的主要是字符的位置在这里无关紧要，所以我们将计算字符的频率，如果它们都相同，那么我们就完成了，不需要删除任何字符来计算频率。字符相同 否则我们可以一个一个地迭代所有字符并将它们的频率减一，如果所有频率都相同，那么我们将标记最多可以通过删除一个字符使字符频率相同，如果频率不匹配然后我们将再次增加该频率并循环其他字符。

以下是上述方法的试运行：

下面是上述方法的实现：

C++

// C++ program to get same frequency character
// string by removal of at most one char
#include 
using namespace std;
#define M 26
 
// Utility method to get index of character ch
// in lower alphabet characters
int getIdx(char ch)
{
    return (ch - 'a');
}
 
// Returns true if all non-zero elements
// values are same
bool allSame(int freq[], int N)
{
    int same;
 
    // get first non-zero element
    int i;
    for (i = 0; i < N; i++) {
        if (freq[i] > 0) {
            same = freq[i];
            break;
        }
    }
 
    // check equality of each element with variable same
    for (int j = i + 1; j < N; j++)
        if (freq[j] > 0 && freq[j] != same)
            return false;
 
    return true;
}
 
// Returns true if we can make all character
// frequencies same
bool possibleSameCharFreqByOneRemoval(string str)
{
    int l = str.length();
 
    // fill frequency array
    int freq[M] = { 0 };
    for (int i = 0; i < l; i++)
        freq[getIdx(str[i])]++;
 
    // if all frequencies are same, then return true
    if (allSame(freq, M))
        return true;
 
    /*  Try decreasing frequency of all character
        by one and then    check all equality of all
        non-zero frequencies */
    for (char c = 'a'; c <= 'z'; c++) {
        int i = getIdx(c);
 
        // Check character only if it occurs in str
        if (freq[i] > 0) {
            freq[i]--;
 
            if (allSame(freq, M))
                return true;
            freq[i]++;
        }
    }
 
    return false;
}
 
// Driver code to test above methods
int main()
{
    string str = "xyyzz";
    if (possibleSameCharFreqByOneRemoval(str))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program to get same frequency character
// string by removal of at most one char
public class GFG {
 
    static final int M = 26;
 
    // Utility method to get index of character ch
    // in lower alphabet characters
    static int getIdx(char ch)
    {
        return (ch - 'a');
    }
 
    // Returns true if all non-zero elements
    // values are same
    static boolean allSame(int freq[], int N)
    {
        int same = 0;
 
        // get first non-zero element
        int i;
        for (i = 0; i < N; i++) {
            if (freq[i] > 0) {
                same = freq[i];
                break;
            }
        }
 
        // check equality of each element with
        // variable same
        for (int j = i + 1; j < N; j++)
            if (freq[j] > 0 && freq[j] != same)
                return false;
 
        return true;
    }
 
    // Returns true if we can make all character
    // frequencies same
    static boolean possibleSameCharFreqByOneRemoval(String str)
    {
        int l = str.length();
 
        // fill frequency array
        int[] freq = new int[M];
 
        for (int i = 0; i < l; i++)
            freq[getIdx(str.charAt(i))]++;
 
        // if all frequencies are same, then return true
        if (allSame(freq, M))
            return true;
 
        /*  Try decreasing frequency of all character
            by one and then check all equality of all
            non-zero frequencies */
        for (char c = 'a'; c <= 'z'; c++) {
            int i = getIdx(c);
 
            // Check character only if it occurs in str
            if (freq[i] > 0) {
                freq[i]--;
 
                if (allSame(freq, M))
                    return true;
                freq[i]++;
            }
        }
 
        return false;
    }
 
    // Driver code to test above methods
    public static void main(String args[])
    {
        String str = "xyyzz";
        if (possibleSameCharFreqByOneRemoval(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to get same frequency character
# string by removal of at most one char
M = 26
 
# Utility method to get index of character ch
# in lower alphabet characters
def getIdx(ch):
    return (ord(ch) - ord('a'))
 
# Returns true if all non-zero elements
# values are same
def allSame(freq, N):
     
    # get first non-zero element
    for i in range(0, N):
        if(freq[i] > 0):
            same = freq[i]
            break
         
    # check equality of each element
    # with variable same    
    for j in range(i + 1, N):
        if(freq[j] > 0 and freq[j] != same):
            return False
 
    return True
 
# Returns true if we can make all
# character frequencies same
def possibleSameCharFreqByOneRemoval(str1):
    l = len(str1)
 
    # fill frequency array
    freq = [0] * M
    for i in range(0, l):
        freq[getIdx(str1[i])] += 1
         
    # if all frequencies are same,
    # then return true
    if(allSame(freq, M)):
        return True
     
    # Try decreasing frequency of all character
    # by one and then check all equality of all
    # non-zero frequencies
    for i in range(0, 26):
         
        # Check character only if it
        # occurs in str
        if(freq[i] > 0):
            freq[i] -= 1
 
            if(allSame(freq, M)):
                return True
            freq[i] += 1
 
    return False
 
# Driver code
if __name__ == "__main__":
    str1 = "xyyzz"
    if(possibleSameCharFreqByOneRemoval(str1)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Sairahul099

C#

// C# program to get same frequency
// character string by removal of
// at most one char
using System;
 
class GFG {
    static int M = 26;
 
    // Utility method to get
    // index of character ch
    // in lower alphabet characters
    static int getIdx(char ch)
    {
        return (ch - 'a');
    }
 
    // Returns true if all
    // non-zero elements
    // values are same
    static bool allSame(int[] freq,
                        int N)
    {
        int same = 0;
 
        // get first non-zero element
        int i;
        for (i = 0; i < N; i++) {
            if (freq[i] > 0) {
                same = freq[i];
                break;
            }
        }
 
        // check equality of
        // each element with
        // variable same
        for (int j = i + 1; j < N; j++)
            if (freq[j] > 0 && freq[j] != same)
                return false;
 
        return true;
    }
 
    // Returns true if we
    // can make all character
    // frequencies same
    static bool possibleSameCharFreqByOneRemoval(string str)
    {
        int l = str.Length;
 
        // fill frequency array
        int[] freq = new int[M];
 
        for (int i = 0; i < l; i++)
            freq[getIdx(str[i])]++;
 
        // if all frequencies are same,
        // then return true
        if (allSame(freq, M))
            return true;
 
        /* Try decreasing frequency of all
            character by one and then check
            all equality of all non-zero
            frequencies */
        for (char c = 'a'; c <= 'z'; c++) {
            int i = getIdx(c);
 
            // Check character only if
            // it occurs in str
            if (freq[i] > 0) {
                freq[i]--;
 
                if (allSame(freq, M))
                    return true;
                freq[i]++;
            }
        }
 
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        string str = "xyyzz";
        if (possibleSameCharFreqByOneRemoval(str))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed
// by ChitraNayal

PHP

 0)
        {
            $same = $freq[$i];
            break;
        }
    }
 
    // check equality of each
    // element with variable same
    for ($j = $i + 1; $j < $N; $j++)
        if ($freq[$j] > 0 &&
            $freq[$j] != $same)
            return false;
 
    return true;
}
 
// Returns true if we
// can make all character
// frequencies same
function possibleSameCharFreqByOneRemoval($str)
{
    global $M;
    $l = strlen($str);
 
    // fill frequency array
    $freq = array_fill(0, $M, NULL);
    for ($i = 0; $i < $l; $i++)
        $freq[getIdx($str[$i])]++;
 
    // if all frequencies are same,
    // then return true
    if (allSame($freq, $M))
        return true;
 
    /* Try decreasing frequency of all
        character by one and then check
        all equality of all non-zero
        frequencies */
    for ($c = 'a'; $c <= 'z'; $c++)
    {
        $i = getIdx($c);
 
        // Check character only
        // if it occurs in str
        if ($freq[$i] > 0)
        {
            $freq[$i]--;
 
            if (allSame($freq, $M))
                return true;
            $freq[$i]++;
        }
    }
 
    return false;
}
 
// Driver code
$str = "xyyzz";
if (possibleSameCharFreqByOneRemoval($str))
echo "Yes";
else
echo "No";
 
// This code is contributed
// by ChitraNayal
?>

Javascript

 
// Javascript program to get same
// frequency character string by
// removal of at most one char
let  M = 26;
 
// Utility method to get index of character
// ch in lower alphabet characters
function getIdx(ch)
{
    return (ch - 'a');
}
 
// Returns true if all non-zero elements
// values are same
function allSame(freq, N)
{
    let same = 0;
 
    // Get first non-zero element
    let i;
    for(i = 0; i < N; i++)
    {
        if (freq[i] > 0)
        {
            same = freq[i];
            break;
        }
    }
 
    // Check equality of each element with
    // variable same
    for(let j = i + 1; j < N; j++)
        if (freq[j] > 0 && freq[j] != same)
            return false;
 
    return true;
}
 
// Returns true if we can make all character
// frequencies same
function possibleSameCharFreqByOneRemoval(str)
{
    let l = str.length;
 
    // Fill frequency array
    let freq = new Array(M);
    for(let i = 0; i < M; i++)
    {
        freq[i] = 0;
    }
 
    for(let i = 0; i < l; i++)
        freq[getIdx(str[i])]++;
 
    // If all frequencies are same,
    // then return true
    if (allSame(freq, M))
        return true;
 
    // Try decreasing frequency of all character
    // by one and then check all equality of all
    // non-zero frequencies
    for(let c = 'a'; c <= 'z'; c++)
    {
        let i = getIdx(c);
 
        // Check character only if
        // it occurs in str
        if (freq[i] > 0)
        {
            freq[i]--;
 
            if (allSame(freq, M))
                return true;
                 
            freq[i]++;
        }
    }
    return false;
}
 
// Driver code
let str = "xyyzz";
 
if (possibleSameCharFreqByOneRemoval(str))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by avanitrachhadiya2155
 

输出：

Yes

时间复杂度： O(n) 假设字母大小是恒定的。

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