Rain water over buildings - Python3 - Analyst test

Rain water over buildings - Python3 - Analyst test.ipynb

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      "source": "**Question:**\n\nIt is rainy season in the city. What vol. of water will be held above adjoining buildings of `h` height, where:\n\n**Assumptions:**\n\n1) Water does not leak to the front or rear of the buildings.\n\n2) Water 'leaks' from all buildings if, at n index, all preceding and subsequent building heights decline sequentially resulting in 0 cubic metres of water being retained above all buildings.\n\n3) At the end of the row of buildings or, technically speaking, above i[0] and i[-1], no water is held due to an absensce of adjoing buildings to both sides."
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      "source": "h =  [9,8,7,8,9,5,6]\n#h1 = [4,5,6,6,6,5,1]\n#h2 = [4,5,8,6,6,8,9,7,6,8,5,1]\n#h3 = [6,5,7,9,9,3,8,6,5,3,7,9,9,3,7,3,6,5,5,3,7,8,9,9,3]\n\n\ndef cubicM(h): \n    \n    l = len(h)\n    hmax = max(h)\n    hr = list(reversed(h))\n    if [x for x in h if h.count(x) > 1]:\n        maxpeaks = max(set([x for x in h if h.count(x) > 1]))\n    else: None\n    firstmaxindex = h.index(maxpeaks)\n    lastmaxindex = (len(h)-1)-hr.index(maxpeaks)\n    peaksrange = h[firstmaxindex:lastmaxindex+1]\n\n    #keep count of cubic metres of water held above building with:\n    \n    cubcount = 0\n    \n    #Logic to calculate the amount of water held over the building in the sub-retaining range:\n    \n    if maxpeaks is not None:\n            for i in peaksrange:\n                if maxpeaks - i > 0:\n                    cubcount+= maxpeaks - i\n            \n    #where a sub-retaining range is found, creating 2 sub lists list from the remaining to the left and right:\n    \n            ltarr = h[:(firstmaxindex)]\n            ltlen = len(ltarr)\n            rtarr = h[lastmaxindex+1:]\n            revrtarr =list(reversed(rtarr))\n            rtlen = len(rtarr)\n            \n    #logic to calculate the amount of water held over buildings before peak range:\n    \n            if ltarr:\n                a = 0\n                while a < ltlen-1:\n                    tmpmax = 0\n                    if ltarr[a] < tmpmax:\n                        cubcount+=ltarr[a] - tmpmax\n                        a+=1\n                    elif ltarr[a] > tmpmax: \n                        tmpmax = ltarr[a]\n                        a+=1\n                    else:\n                        a+=1\n                        \n    #logic to calculate the amount of water held over buildings after peak range:\n    \n            elif rtarr:\n                b = 0\n                tmpmax = revrtarr[0]\n                while b < rtlen:\n                    if revrtarr[b] < tmpmax:\n                        cubcount+=tmpmax - revrtarr[b]\n                        b+=1\n                    elif rtarr[b] > tmpmax: \n                        tmpmax = revrtarr[b]\n                        b+=1\n                    else:\n                        b+=1 \n                        \n    #where no sub-retaining range is found in list, at the tallest building create sub list to the left: \n    \n    elif maxpeaks is None:\n            x = 0 \n            arr1 = h[0:h.index(hmax)]\n            tmpmax = arr1[0]\n            while x < len(arr1)-1:\n                if arr1[x] < tmpmax:\n                    cubcount+=arr1[x] - tmpmax\n                    x+=1\n                elif arr1[x] > tmpmax: \n                    tmpmax = arr1[x]\n                    x+=1\n                else:\n                    x+=1           \n    #...and to the right:\n    \n            arr2 = h[h.index(hmax):]\n            revarr2 = list(reversed(arr2))\n            tmpmax = revarr2[0]\n            while x in len(revarr2) -1:\n                if revarr2[x] < tmpmax:\n                    cubcount+=revarr2[x] - tmpmax\n                    x+=1\n                elif revarr2[x] > tmpmax: \n                    tmpmax = revarr2[x]\n                    x+=1\n                else:\n                    x+=1\n    return(cubcount)\n\nprint(cubicM(h))",
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