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How to use oneleft method in pyatom

Best Python code snippet using pyatom_python

PowerBinaryCalculator.py

Source:PowerBinaryCalculator.py

1#generate inputs 2input = open('input.txt','r')3numbers = input.read().splitlines()4cNum= numbers.copy()5#values used in script6lenLine=len(numbers[0])7sigBit,oneLeft="0",false8#calculate oxygen9#for every bit position(0->11 in this case)10for bitPos in range(0,lenLine):11    #count the number of 0s and 1s 12    zeros,ones=0,013    for i in range(0,len(numbers)):14        if((numbers[i])[bitPos]=="0"):15            zeros+=116        else:17            ones+=118    #if ones>=zeros:19    if(ones>=zeros):20        sigBit="1"21    else:22        sigBit="0"23    #remove any items in the list that don't meet oxygen specifications24    i=025    while(i<len(numbers)):26        if(len(oxNum)==1):27            oneLeft=true28            break29        if((numbers[i])[bitPos]!=sigBit):30            numbers.pop(i)31        else:32            i+=133    if(oneLeft): #stop if only one result left34        break35#calculate c0236#for every bit position(0->11 in this case)37oneLeft=false38for bitPos in range(0,lenLine):39    #count the number of 0s and 1s 40    zeros,ones=0,041    for i in range(0,len(cNum)):42        if((cNum[i])[bitPos]=="0"):43            zeros+=144        else:45            ones+=146    #if ones>=zeros:(inverse of oxygen)47    if(ones>=zeros):48        sigBit="0"49    else:50        sigBit="1"51    #remove any items in the list that don't meet oxygen specifications52    i=053    while(i<len(cNum)):54        if(len(cNum)==1):55            oneLeft=true56            break57        if((cNum[i])[bitPos]!=sigBit):58            cNum.pop(i)59        else:60            i+=161    if(oneLeft): #stop if only one result left62        break63#generate and print results64oxygen=int(numbers[0],2)65co2=int(cNum[0],2)...

postfixlogic.py

Source:postfixlogic.py

1__author__ = "Dmitry Philippov"2import sys3sys.stdout = open("postfixlogic.out", "w")4def main():5    print("""16S 0 -> now 0 >7S 1 -> now 1 >8now _ -> fillblank _ <9fillblank * -> fillblank _ <10fillblank 0 -> AC 0 ^11fillblank 1 -> AC 1 ^12now 0 -> zeroleft * <13zeroleft _ -> setzero _ >14zeroleft * -> zeroleft * <15zeroleft _ -> now _ ^16zeroleft 0 -> setzero 0 >17zeroleft 1 -> setzero 1 >18setzero * -> zeroright 0 >19zeroright * -> zeroright * >20zeroright _ -> now _ ^21zeroright 0 -> now 0 ^22zeroright 1 -> now 1 ^23zeroright o -> now o ^24zeroright a -> now a ^25now 1 -> oneleft * <26oneleft _ -> setone _ >27oneleft * -> oneleft * <28oneleft 0 -> setone 0 >29oneleft 1 -> setone 1 >30setone * -> oneright 1 >31oneright * -> oneright * >32oneright 0 -> now 0 ^33oneright 1 -> now 1 ^34oneright o -> now o ^35oneright a -> now a ^36now a -> andleft * <37andleft * -> andleft * <38andleft 0 -> andleftzero * <39andleftzero 0 -> andright 0 >40andleftzero 1 -> andright 0 >41andright * -> andright * >42andright _ -> now _ ^43andright 0 -> now 0 ^44andright 1 -> now 1 ^45andright o -> now o ^46andright a -> now a ^47andleft 1 -> andleftone * <48andleftone 0 -> andright 0 >49andleftone 1 -> andright 1 >50now o -> orleft * <51orleft * -> orleft * <52orleft 0 -> orleftzero * <53orleftzero 0 -> orright 0 >54orleftzero 1 -> orright 1 >55orright * -> orright * >56orright _ -> now _ ^57orright 0 -> now 0 ^58orright 1 -> now 1 ^59orright o -> now o ^60orright a -> now a ^61orleft 1 -> orleftone * <62orleftone 0 -> orright 1 >63orleftone 1 -> orright 1 >""")64if __name__ == "__main__":...

034-median-of-two-sorted-arrays.py

Source:034-median-of-two-sorted-arrays.py

1class Solution:2    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:3        m, n = len(nums1), len(nums2)4        halfLen = (m + n) // 25        6        if m > n:7            nums1, nums2 = nums2, nums18            m, n = n, m9            10        l, r = 0, m - 111        while True:12            i = l + (r - l) // 213            j = halfLen - i - 214            15            oneLeft = nums1[i] if i >= 0 else float('-infinity')16            oneRight = nums1[i + 1] if (i + 1) < m else float('infinity')17            twoLeft = nums2[j] if j >= 0 else float('-infinity')18            twoRight = nums2[j + 1] if (j + 1) < n else float('infinity')19            20            if oneLeft <= twoRight and twoLeft <= oneRight:21                if (m + n) % 2:22                    return min(oneRight, twoRight)23                return (min(oneRight, twoRight) + max(oneLeft, twoLeft)) / 224            elif oneLeft > twoRight:25                r = i - 126            else:...

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