I have a N*N Matrix.Now i want to know the diagonal difference of this Matrix.What will be the best approach of this solution?
I am trying with given approach:
Such as it is 3*3 Matrix say it is:
11 15 85
66 72 21
14 21 47
the diagonal simple formula will be:
firstD= (11+72+47) = 130
secondD = (85+72+14)= 171
diagonalDiff = |firstD - secondD| = |130-171| = 41
If I count every row such as first to find out firstD (First row's first value + Sec row's Sec value + Third row's third value+..).This is my thinking. Can anyone tell me best approaches?
Try this:
$arr = array(
array(11, 15, 85),
array(66, 72, 21),
array(14, 21, 47),
);
$arrDiag = count($arr);
$firstD = 0;
$secondD = 0;
$i = 0;
for($j = 0; $j < $arrDiag; $j++){
$firstD += $arr[$i++][$j];
$secondD += $arr[$arrDiag - $i][$j];
}
echo abs($firstD - $secondD);//41
Here is a pseudo code for your problem using one simple loop:
// $array - predefined 2 dimentional array with $N rows and $N columns
$result = 0;
for ($i=0;$i<$N;$i++)
{
$result += ($array[$i,$i] - &array[$i,$N-$i-1]);
}
return echo abs($result);
that way you can do the calculation in one pass, and do a diff between two elements in each row insead of calculation the sum of each diagonal
Model your matrix with a multi-dimensional array and iterate through it. The easiest way should be the following:
<?
$matrix = array(array(1,2,3),array(4,5,6),array(7,8,9)); //Insert or define your matrix here..
$n = count($matrix); //Size of matrix, thanks to VolkerK
$firstD = 0;
$lastD = 0;
for($i = 0; $i < $n; $i++){
$firstD += $matrix[$i][$i];
$lastD += $matrix[$i][$n-$i-1];
}
echo $firstD."
";
echo $lastD;
This is the code you need:
$first = 0;
$second = 0;
for($i = 0; $i < N; $i++) {
for($j = 0; $j < N; $j++) {
if($i == $j) {
$first += $matrix[$i][$j];
} else if($i + $j == N) {
$second += $matrix[$i][$j];
}
}
}
$diagonalDiff = abs($first - $second);
Where $matrix is a N*N array
you can try this :
$first_diag=$second_diag=0;
$matrix=array(array(11,15,85),array(66,72,21),array(14,21,47));
foreach($matrix as $index=>$sub_array){
$first_diag +=$sub_array[$index];
$second_diag +=$sub_array[count($matrix)-1-$index];
}
print abs ($first_diag-$second_diag);
For one you can use a matrix library like Math_Matrix. But if this is the only operation you are gona need then it's best to write a generalized function for this using the same method you quoted yourself.
function diagonalDiff($n){
$firstD = 0;
$secondD = 0;
for($i=0;$i<$n;$i++){
for($j=0;$j<$n;$j++){
if($i == $j) {
$first += $matrix[$i][$j];
} else if($i + $j == $n) {
$secondD += $matrix[$i][$j];
}
}
}
return abs($firstD-$secondD);
}
Should give you the answer you need for a matrix of a given size $n. (Only square Matrixes ofcourse :) )
Another better solution and it's easy to understand:
<?php
$arr = array(
array(11, 15, 85),
array(66, 72, 21),
array(14, 21, 47),
);
$arrDiag = count($arr);
$firstD = 0;
$secondD = 0;
$i = 0;
for($j = 0; $j < $arrDiag; $j++){
$i++;
$firstD += $arr[$j][$j];
$secondD += $arr[$arrDiag - $i][$j];
}
echo abs($firstD - $secondD);
?>
Recently solved this question in Hacker Rank.
Just using the function array_reduce:
function diagonalDifference($arr) {
$i = 0;
$n = count($arr);
return abs(array_reduce($arr,
function ($c, $str) use (&$i, $n ) {
$i++;
return $c + $str[$i-1] - $str[$n-$i];
}, 0));
}
<?php
$m=array(array(3,4,-1,11,2),
array(-3,2,1,6,9),
array(3,4,6,5,-2),
array(1,9,9,7,-3),
array(12,9,16,7,-3));
echo count($m)."<br>";
$i=0;
$j=0;
$ek=0;
$k=0;
$n=1;
$es=count($m)-1;
for($i=0;$i<count($m);$i++)
{
for($j=0;$j<count($m);$j++)
{
echo $m[$i][$j]." ";
}
echo "<br>";
}
echo "<br>";
for($k=0;$k<count($m);$k++)
{
for($n=0;$n<count($m);$n++)
{
if($k==$n){
$ek=$ek+$m[$k][$n];
}
}
echo "<br>";
}
echo "<hr>";
$p=0;
for($k=0;$k<count($m);$k++)
{
echo $m[$k][$es-$p]."<br> ";
$p++;
}
?>
Try this:
function diagonalDifference($arr) {
$left = 0;
$right = 0;
$i = 0;
foreach($arr as $ar){
$left+= $ar[0+$i];
$right+= $ar[count($ar) - (1+$i)];
$i++;
}
return abs($left - $right);
}