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codeparrot

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xlcost-text-to-code

like 35

In worst case it will be arr [ i - 1 ] + 1 we use all combination of a [ i - 1 ] and add 1

arr [ i ] = arr [ i - 1 ] + 1 NEW_LINE sqrNum [ i ] = 1 NEW_LINE k = 1 ; NEW_LINE

Check for all square number less or equal to i

while ( k * k <= i ) : NEW_LINE

If it gives less count then update it

if ( arr [ i ] > arr [ i - k * k ] + 1 ) : NEW_LINE INDENT arr [ i ] = arr [ i - k * k ] + 1 NEW_LINE sqrNum [ i ] = k * k NEW_LINE DEDENT k += 1 NEW_LINE

v stores optimum square number whose sum give N

while ( n > 0 ) : NEW_LINE INDENT v . append ( sqrNum [ n ] ) NEW_LINE n -= sqrNum [ n ] ; NEW_LINE DEDENT return v NEW_LINE

Driver code

n = 10 NEW_LINE

Calling function

v = minSqrNum ( n ) NEW_LINE

Printing vector

for i in range ( len ( v ) ) : NEW_LINE INDENT print ( v [ i ] , end = " " ) NEW_LINE if ( i < len ( v ) - 1 ) : NEW_LINE INDENT print ( " ▁ + ▁ " , end = " " ) NEW_LINE DEDENT DEDENT