原文:https://www . geesforgeks . org/minimum-sum-two-elements-two-array-indexes-not/
给定两个大小相同的数组 a[]和 b[]。任务是找到两个元素的最小和，使它们属于不同的数组，并且不在数组的同一索引处。
示例:
Input : a[] = {5, 4, 13, 2, 1}
b[] = {2, 3, 4, 6, 5}
Output : 3
We take 1 from a[] and 2 from b[]
Sum is 1 + 2 = 3.
Input : a[] = {5, 4, 13, 1}
b[] = {3, 2, 6, 1}
Output : 3
We take 1 from a[] and 2 from b[].
Note that we can't take 1 from b[]
as the elements can not be at same
index.
A 简单的解决方案是考虑 a[]的每个元素，在不同于其索引的索引处与 b[]的所有元素形成对，并计算和。最后返回最小和。该解决方案的时间复杂度为 0(n2
一个高效的解决方案在 O(n)时间内起作用。以下是步骤。
以下是上述想法的实现:
// C++ program to find minimum sum of two
// elements chosen from two arrays such that
// they are not at same index.
#include
using namespace std;
// Function which returns minimum sum of two
// array elements such that their indexes are
// not same
int minSum(int a[], int b[], int n)
{
// Finding minimum element in array A and
// also/ storing its index value.
int minA = a[0], indexA;
for (int i=1; i
if (a[i]
minA = a[i];
indexA = i;
}
}
// Finding minimum element in array B and
// also storing its index value
int minB = b[0], indexB;
for (int i=1; i
if (b[i]
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements are
// not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as previous
// and value is also less than other minimum
// Store new minimum and store its index
int minA2 = INT_MAX, indexA2;
for (int i=0; i
if (i != indexA && a[i]
minA2 = a[i];
indexA2 = i;
}
}
// When index of B is not same as previous
// and value is also less than other minimum.
// Store new minimum and store its index
int minB2 = INT_MAX, indexB2;
for (int i=0; i
if (i != indexB && b[i]
minB2 = b[i];
indexB2 = i;
}
}
// Taking sum of previous minimum of a[]
// with new minimum of b[]
// and also sum of previous minimum of b[]
// with new minimum of a[]
// and return whichever is minimum.
return min(minB + minA2, minA + minB2);
}
// Driver code
int main()
{
int a[] = {5, 4, 3, 8, 1};
int b[] = {2, 3, 4, 2, 1};
int n = sizeof(a)/sizeof(a[0]);
cout <
}
// Java program to find minimum sum of two
// elements chosen from two arrays such that
// they are not at same index.
class Minimum{
// Function which returns minimum sum of two
// array elements such that their indexes are
// not same
public static int minSum(int a[], int b[], int n)
{
// Finding minimum element in array A and
// also/ storing its index value.
int minA = a[0], indexA = 0;
for (int i=1; i
if (a[i]
minA = a[i];
indexA = i;
}
}
// Finding minimum element in array B and
// also storing its index value
int minB = b[0], indexB = 0;
for (int i=1; i
if (b[i]
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements are
// not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as previous
// and value is also less than other minimum
// Store new minimum and store its index
int minA2 = Integer.MAX_VALUE, indexA2 = 0;
for (int i=0; i
if (i != indexA && a[i]
minA2 = a[i];
indexA2 = i;
}
}
// When index of B is not same as previous
// and value is also less than other minimum.
// Store new minimum and store its index
int minB2 = Integer.MAX_VALUE, indexB2 = 0;
for (int i=0; i
if (i != indexB && b[i]
minB2 = b[i];
indexB2 = i;
}
}
// Taking sum of previous minimum of a[]
// with new minimum of b[]
// and also sum of previous minimum of b[]
// with new minimum of a[]
// and return whichever is minimum.
return Math.min(minB + minA2, minA + minB2);
}
public static void main(String[] args)
{
int a[] = {5, 4, 3, 8, 1};
int b[] = {2, 3, 4, 2, 1};
int n = 5;
System.out.print(minSum(a, b, n));
}
}
// This code is contributed by rishabh_jain
# Python3 code to find minimum sum of
# two elements chosen from two arrays
# such that they are not at same index.
import sys
# Function which returns minimum sum
# of two array elements such that their
# indexes arenot same
def minSum(a, b, n):
# Finding minimum element in array A
# and also storing its index value.
minA = a[0]
indexA = 0
for i in range(1,n):
if a[i]
indexA = i
# Finding minimum element in array B
# and also storing its index value
minB = b[0]
indexB = 0
for i in range(1, n):
if b[i]
indexB = i
# If indexes of minimum elements
# are not same, return their sum.
if indexA != indexB:
return (minA + minB)
# When index of A is not same as
# previous and value is also less
# than other minimum. Store new
# minimum and store its index
minA2 = sys.maxsize
indexA2=0
for i in range(n):
if i != indexA and a[i]
indexA2 = i
# When index of B is not same as
# previous and value is also less
# than other minimum. Store new
# minimum and store its index
minB2 = sys.maxsize
indexB2 = 0
for i in range(n):
if i != indexB and b[i]
indexB2 = i
# Taking sum of previous minimum of
# a[] with new minimum of b[]
# and also sum of previous minimum
# of b[] with new minimum of a[]
# and return whichever is minimum.
return min(minB + minA2, minA + minB2)
# Driver code
a = [5, 4, 3, 8, 1]
b = [2, 3, 4, 2, 1]
n = len(a)
print(minSum(a, b, n))
# This code is contributed by "Sharad_Bhardwaj".
// C# program to find minimum sum of
// two elements chosen from two arrays
// such that they are not at same index.
using System;
public class GFG {
// Function which returns minimum
// sum of two array elements such
// that their indexes are not same
static int minSum(int []a, int []b,
int n)
{
// Finding minimum element in
// array A and also/ storing its
// index value.
int minA = a[0], indexA = 0;
for (int i = 1; i
if (a[i]
minA = a[i];
indexA = i;
}
}
// Finding minimum element in
// array B and also storing its
// index value
int minB = b[0], indexB = 0;
for (int i = 1; i
if (b[i]
minB = b[i];
indexB = i;
}
}
// If indexes of minimum elements
// are not same, return their sum.
if (indexA != indexB)
return (minA + minB);
// When index of A is not same as
// previous and value is also less
// than other minimum Store new
// minimum and store its index
int minA2 = int.MaxValue;
for (int i=0; i
if (i != indexA && a[i]
minA2 = a[i];
}
}
// When index of B is not same as
// previous and value is also less
// than other minimum. Store new
// minimum and store its index
int minB2 = int.MaxValue;
for (int i=0; i
// Taking sum of previous minimum
// of a[] with new minimum of b[]
// and also sum of previous minimum
// of b[] with new minimum of a[]
// and return whichever is minimum.
return Math.Min(minB + minA2,
minA + minB2);
}
public static void Main()
{
int []a = {5, 4, 3, 8, 1};
int []b = {2, 3, 4, 2, 1};
int n = 5;
Console.Write(minSum(a, b, n));
}
}
// This code is contributed by Sam007.
// PHP program to find minimum
// sum of two elements chosen
// from two arrays such that
// they are not at same index.
// Function which returns
// minimum sum of two array
// elements such that their
// indexes are not same
function minSum($a, $b, $n)
{
// Finding minimum element
// in array A and also
// storing its index value.
$minA = $a[0];
for ($i = 1; $i <$n; $i++)
{
if ($a[$i] <$minA)
{
$minA = $a[$i];
$indexA = $i;
}
}
// Finding minimum element
// in array B and also
// storing its index value
$minB = $b[0];
for ($i = 1; $i <$n; $i++)
{
if ($b[$i] <$minB)
{
$minB = $b[$i];
$indexB = $i;
}
}
// If indexes of minimum
// elements are not same,
// return their sum.
if ($indexA != $indexB)
return ($minA + $minB);
// When index of A is not
// same as previous and
// value is also less than
// other minimum. Store new
// minimum and store its index
$minA2 = 9999999;
$indexA2 = 0;
for ($i = 0; $i <$n; $i++)
{
if ($i != $indexA &&
$a[$i] <$minA2)
{
$minA2 = $a[$i];
$indexA2 = $i;
}
}
// When index of B is not
// same as previous and
// value is also less than
// other minimum. Store new
// minimum and store its index
$minB2 = 999999;
$indexB2 = 0;
for ($i = 0; $i <$n; $i++)
{
if ($i != $indexB &&
$b[$i] <$minB2)
{
$minB2 = $b[$i];
$indexB2 = $i;
}
}
// Taking sum of previous
// minimum of a[] with
// new minimum of b[]
// and also sum of previous
// minimum of b[] with new
// minimum of a[]
// and return whichever
// is minimum.
return min($minB + $minA2,
$minA + $minB2);
}
// Driver code
$a = array(5, 4, 3, 8, 1);
$b = array(2, 3, 4, 2, 1);
$n = count($a);
echo minSum($a, $b, $n);
// This code is contributed
// by Sam007
?>
输出:
3
时间复杂度:O(n)
T3】辅助空间: O(1)
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