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Minimum sum of absolute differences of pairs in a triplet from three arrays | Lower_bound function.
Function to find the value closest to K in the array A [ ].
Initialize close value as the end element.
Find lower bound of the array.
If lower_bound is found.
If lower_bound is not the first array element.
If * ( it - 1 ) is closer to k.
Return closest value of k.
Function to find the minimum sum of abs ( arr [ i ] - brr [ j ] ) and abs ( brr [ j ] - crr [ k ] ).
Sort the vectors arr and crr.
Initialize minimum as LET_MAX.
Traverse the array brr [ ].
Stores the element closest to val from the array arr [ ].
Stores the element closest to val from the array crr [ ].
If sum of differences is minimum.
Update the minimum.
Print the minimum absolute difference possible.
Driver code.
Function Call"
Solution in Python:

Write me a code to minimum sum of absolute differences of pairs in a triplet from three arrays and the other steps required?
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#Here is code to minimum sum of absolute differences of pairs in a triplet from three arrays and the other steps required.
 def lower_bound ( arr , key ) :
	low = 0 ;
	high = len ( arr ) - 1 ;
	while ( low < high ) :
		mid = low + ( high - low ) // 2 ;
		if ( arr [ mid ] >= key ) :
			high = mid ;
		else :
			low = mid + 1 ;
	return low ;
def closestValue ( A , k ) :
	close = A [ - 1 ] ;
	it = lower_bound ( A , k ) ;
	if ( it != len ( A ) ) :
		close = A [ it ] ;
		if ( it != 0 ) :
			if ( ( k - A [ it - 1 ] ) < ( close - k ) ) :
				close = A [ it - 1 ] ;
	return close ;
def minPossible ( arr , brr , crr ) :
	arr . sort ( ) ;
	crr . sort ( ) ;
	minimum = 10 ** 9 ;
	for val in brr :
		arr_close = closestValue ( arr , val ) ;
		crr_close = closestValue ( crr , val ) ;
		if ( abs ( val - arr_close ) + abs ( val - crr_close ) < minimum ) :
			minimum = abs ( val - arr_close ) + abs ( val - crr_close ) ;
	print ( minimum ) ;
a = [ 1 , 8 , 5 ] ;
b = [ 2 , 9 ] ;
c = [ 5 , 4 ] ;
minPossible ( a , b , c ) ;