满足给定条件的字符串数量
给定相等长度的 N 根弦。字符串只包含数字(1 到 9)。任务是计算具有索引位置的字符串的数量，使得该索引位置的数字大于所有其他字符串的相同索引位置的数字。 例:
输入:arr[]= {“223”、“232”、“112”} 输出:2 1ST和 2 nd 的第一位数字最大。 字符串 2 的第二个数字和最大。 字符串 1 的第三位数字 st 最大。 输入:arr[]= {“999”、“122”、“111”} 输出: 1
方法:对于每个索引位置，在所有字符串中找到该位置的最大数字。并将满足给定条件的字符串的索引存储在集合中，以便同一字符串不会在不同的索引位置被计数两次。最后，返回集合的大小。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of valid strings
int countStrings(int n, int m, string s[])
{
// Set to store indices of valid strings
unordered_set<int> ind;
for (int j = 0; j < m; j++) {
int mx = 0;
// Find the maximum digit for current position
for (int i = 0; i < n; i++)
mx = max(mx, (int)s[i][j] - '0');
// Add indices of all the strings in the set
// that contain maximal digit
for (int i = 0; i < n; i++)
if (s[i][j] - '0' == mx)
ind.insert(i);
}
// Return number of strings in the set
return ind.size();
}
// Driver code
int main()
{
string s[] = { "223", "232", "112" };
int m = s[0].length();
int n = sizeof(s) / sizeof(s[0]);
cout << countStrings(n, m, s);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GfG
{
// Function to return the count of valid strings
static int countStrings(int n, int m, String s[])
{
// Set to store indices of valid strings
HashSet<Integer> ind = new HashSet<Integer>();
for (int j = 0; j < m; j++)
{
int mx = 0;
// Find the maximum digit for current position
for (int i = 0; i < n; i++)
mx = Math.max(mx, (int)(s[i].charAt(j) - '0'));
// Add indices of all the strings in the set
// that contain maximal digit
for (int i = 0; i < n; i++)
if (s[i].charAt(j) - '0' == mx)
ind.add(i);
}
// Return number of strings in the set
return ind.size();
}
// Driver code
public static void main(String[] args)
{
String s[] = { "223", "232", "112" };
int m = s[0].length();
int n = s.length;
System.out.println(countStrings(n, m, s));
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# Function to return the count of
# valid strings
def countStrings(n, m, s):
# Set to store indices of
# valid strings
ind = dict()
for j in range(m):
mx = 0
str1 = s[j]
# Find the maximum digit for
# current position
for i in range(n):
mx = max(mx, int(str1[i]))
# Add indices of all the strings in
# the set that contain maximal digit
for i in range(n):
if int(str1[i]) == mx:
ind[i] = 1
# Return number of strings
# in the set
return len(ind)
# Driver code
s = ["223", "232", "112"]
m = len(s[0])
n = len(s)
print(countStrings(n, m, s))
# This code is contributed
# by Mohit Kumar
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GfG
{
// Function to return the count of valid strings
static int countStrings(int n, int m, String[] s)
{
// Set to store indices of valid strings
HashSet<int> ind = new HashSet<int>();
for (int j = 0; j < m; j++)
{
int mx = 0;
// Find the maximum digit for current position
for (int i = 0; i < n; i++)
mx = Math.Max(mx, (int)(s[i][j] - '0'));
// Add indices of all the strings in the set
// that contain maximal digit
for (int i = 0; i < n; i++)
if (s[i][j] - '0' == mx)
ind.Add(i);
}
// Return number of strings in the set
return ind.Count;
}
// Driver code
public static void Main()
{
String []s = { "223", "232", "112" };
int m = s[0].Length;
int n = s.Length;
Console.WriteLine(countStrings(n, m, s));
}
}
/* This code contributed by PrinciRaj1992 */
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the count of valid strings
function countStrings(n,m,s)
{
// Set to store indices of valid strings
let ind = new Set();
for (let j = 0; j < m; j++)
{
let mx = 0;
// Find the maximum digit for current position
for (let i = 0; i < n; i++)
mx = Math.max(mx, (s[i][j].charCodeAt(0) -
'0'.charCodeAt(0)));
// Add indices of all the strings in the set
// that contain maximal digit
for (let i = 0; i < n; i++)
if (s[i][j].charCodeAt(0) - '0'.charCodeAt(0) == mx)
ind.add(i);
}
// Return number of strings in the set
return ind.size;
}
// Driver code
let s=[ "223", "232", "112"];
let m = s[0].length;
let n = s.length;
document.write(countStrings(n, m, s));
// This code is contributed by patel2127
</script>
Output:
2
时间复杂度: O(N * M)，其中 N 为字符串的个数，M 为字符串的长度。
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