最小索引i，以使从索引i到给定索引的所有元素都相等

给定一个整数数组arr []和一个整数pos ，任务是找到最小索引i ，以使从索引i到索引pos的所有元素都相等。

例子：

Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 3 
Output: 1 
Elements in index range [1, 3] are all equal to 1.
Input: arr[] = {2, 1, 1, 1, 5, 2}, pos = 5 
Output: 5 
 

简单方法：从索引pos – 1开始，反向遍历数组，对于第一个索引i ，以使arr [i]！= arr [pos]打印i + 1 ，这是所需的索引。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum required index
int minIndex(int arr[], int n, int pos)
{
    int num = arr[pos];
 
    // Start from arr[pos - 1]
    int i = pos - 1;
    while (i >= 0) {
        if (arr[i] != num)
            break;
        i--;
    }
 
    // All elements are equal
    // from arr[i + 1] to arr[pos]
    return i + 1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 1, 1, 1, 5, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int pos = 4;
     
      // Function Call
    cout << minIndex(arr, n, pos);
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the minimum required index
    static int minIndex(int arr[], int n, int pos)
    {
        int num = arr[pos];
 
        // Start from arr[pos - 1]
        int i = pos - 1;
        while (i >= 0) {
            if (arr[i] != num)
                break;
            i--;
        }
 
        // All elements are equal
        // from arr[i + 1] to arr[pos]
        return i + 1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 1, 1, 1, 5, 2 };
        int n = arr.length;
        int pos = 4;
         
          // Function Call
        System.out.println(minIndex(arr, n, pos));
    }
}
 
// This code is contributed by Code_Mech.

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# required index
def minIndex(arr, n, pos):
 
    num = arr[pos]
 
    # Start from arr[pos - 1]
    i = pos - 1
    while (i >= 0):
        if (arr[i] != num):
            break
        i -= 1
     
    # All elements are equal
    # from arr[i + 1] to arr[pos]
    return i + 1
 
# Driver code
arr = [2, 1, 1, 1, 5, 2 ]
n = len(arr)
pos = 4
 
# Function Call
print(minIndex(arr, n, pos))
 
# This code is contributed by
# Mohit Kumar 29

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the minimum required index
    static int minIndex(int[] arr, int n, int pos)
    {
        int num = arr[pos];
 
        // Start from arr[pos - 1]
        int i = pos - 1;
        while (i >= 0) {
            if (arr[i] != num)
                break;
            i--;
        }
 
        // All elements are equal
        // from arr[i + 1] to arr[pos]
        return i + 1;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 1, 1, 1, 5, 2 };
        int n = arr.Length;
        int pos = 4;
         
          // Function Call
        Console.WriteLine(minIndex(arr, n, pos));
    }
}
 
// This code is contributed
// by Akanksha Rai

PHP

= 0)
    {
        if ($arr[$i] != $num)
            break;
        $i--;
    }
 
    // All elements are equal
    // from arr[i + 1] to arr[pos]
    return $i + 1;
}
 
// Driver code
$arr = array(2, 1, 1, 1, 5, 2 );
$n = sizeof($arr);
$pos = 4;
 
echo minIndex($arr, $n, $pos);
 
// This code is contributed by Ryuga
?>

Javascript

 
// Javascript implementation of the approach
 
// Function to return the minimum required index
function minIndex(arr, n, pos)
{
    var num = arr[pos];
 
    // Start from arr[pos - 1]
    var i = pos - 1;
    while (i >= 0) {
        if (arr[i] != num)
            break;
        i--;
    }
 
    // All elements are equal
    // from arr[i + 1] to arr[pos]
    return i + 1;
}
 
// Driver code
var arr = [ 2, 1, 1, 1, 5, 2 ];
var n = arr.length;
var pos = 4;
 
// Function Call
document.write(minIndex(arr, n, pos));
 
 
// This code is contributed by rrrtnx.

输出

4

时间复杂度： O(N)
空间复杂度： O(1)

高效的方法：

在子数组[0，pos-1]中进行二进制搜索。停止条件为arr [mid] == arr [pos] && arr [mid-1]！= arr [pos]。左移或右移分别取决于arr [mid] == arr [pos]。

执行：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum required index
int minIndex(int arr[], int pos)
{
    int low = 0;
    int high = pos;
    int i = pos;
 
    while (low < high) {
        int mid = (low + high) / 2;
        if (arr[mid] != arr[pos]) {
            low = mid + 1;
        }
        else {
            high = mid - 1;
            i = mid;
            if (mid > 0 && arr[mid - 1] != arr[pos]) {
 
                // Short-cicuit more comparisions as found
                // the border point
                break;
            }
        }
    }
 
    // For cases were high = low + 1 and arr[high] will
    // match with
    // arr[pos] but not arr[low] or arr[mid]. In such
    // iteration the if condition will satisfy and loop will
    // break post that low will be updated. Hence i will not
    // point to the correct index.
    return arr[low] == arr[pos] ? low : i;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 1, 1, 1, 5, 2 };
 
    cout << minIndex(arr, 2) << endl; // Should be 1
    cout << minIndex(arr, 3) << endl; // Should be 1
    cout << minIndex(arr, 4) << endl; // Should be 4
    return 0;
}
 
// This code is contributed by
// anshbikram

Java

// Java implementation of the approach
 
class GFG {
     
      // Function to return the minimum required index
    static int minIndex(int arr[], int pos)
    {
        int low = 0;
        int high = pos;
        int i = pos;
 
        while (low < high) {
            int mid = (low + high) / 2;
            if (arr[mid] != arr[pos]) {
                low = mid + 1;
            }
            else {
                high = mid - 1;
                i = mid;
                if (mid > 0 && arr[mid - 1] != arr[pos]) {
                     
                      // Short-cicuit more comparisions as
                    // found the border point
                    break;
                }
            }
        }
 
        // For cases were high = low + 1 and arr[high] will
        // match with arr[pos] but not arr[low] or arr[mid].
        // In such iteration the if condition will satisfy
        // and loop will break post that low will be
        // updated. Hence i will not point to the correct
        // index.
        return arr[low] == arr[pos] ? low : i;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 1, 1, 1, 5, 2 };
 
        System.out.println(minIndex(arr, 2)); // Should be 1
        System.out.println(minIndex(arr, 3)); // Should be 1
        System.out.println(minIndex(arr, 4)); // Should be 4
    }
}
 
// This code is contributed by
// anshbikram

Python3

# Python3 implementation of the approach
 
# Function to return the minimum
# required index
 
def minIndex(arr, pos):
    low = 0
    high = pos
    i = pos
 
    while low < high:
        mid = (low + high)//2
        if arr[mid] != arr[pos]:
            low = mid + 1
        else:
            high = mid - 1
            i = mid
            if mid > 0 and arr[mid-1] != arr[pos]:
 
                # Short-cicuit more comparisions as found the border point
                break
 
    # For cases were high = low + 1 and arr[high] will match with
    # arr[pos] but not arr[low] or arr[mid]. In such iteration
    # the if condition will satisfy and loop will break post that
    # low will be updated. Hence i will not point to the correct index.
    return low if arr[low] == arr[pos] else i
 
 
# Driver code
arr = [2, 1, 1, 1, 5, 2]
 
print(minIndex(arr, 2))  # Should be 1
print(minIndex(arr, 3))  # Should be 1
print(minIndex(arr, 4))  # Should be 4
 
# This code is contributed by
# anshbikram

C＃

// C# implementation of the approach
using System;
 
class GFG{
     
// Function to return the minimum
// required index
static int minIndex(int []arr, int pos)
{
    int low = 0;
    int high = pos;
    int i = pos;
 
    while (low < high)
    {
        int mid = (low + high) / 2;
        if (arr[mid] != arr[pos])
        {
            low = mid + 1;
        }
        else
        {
            high = mid - 1;
            i = mid;
            if (mid > 0 && arr[mid - 1] != arr[pos])
            {
                 
                // Short-cicuit more comparisions as
                // found the border point
                break;
            }
        }
    }
 
    // For cases were high = low + 1 and arr[high] will
    // match with arr[pos] but not arr[low] or arr[mid].
    // In such iteration the if condition will satisfy
    // and loop will break post that low will be
    // updated. Hence i will not point to the correct
    // index.
    return arr[low] == arr[pos] ? low : i;
}
 
// Driver code
public static void Main()
{
    int []arr = { 2, 1, 1, 1, 5, 2 };
 
    Console.WriteLine(minIndex(arr, 2)); // Should be 1
    Console.WriteLine(minIndex(arr, 3)); // Should be 1
    Console.WriteLine(minIndex(arr, 4)); // Should be 4
}
}
 
// This code is contributed by chitranayal

输出

1
1
4

时间复杂度： O(log(n))
空间复杂度： O(1)