The worst solution I've ever written (this is horrible).
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public static final double NATURAL_LOG_10 = Math.log(10);\n    public static final double NATURAL_LOG_2 = Math.log(2);\n    public static final double NATURAL_LOG_17 = Math.log(17);\n    public static final double NATURAL_LOG_3 = Math.log(3);\n    double ERR = 0.00000012;\n    public String abbreviateProduct(int left, int right) {\n        //for 0s, find number of 5 and 2 factors.\n        //for last 5 digits, after taking out 5 and 2 factors, the product mod 100000 should get the last five digits\n        //log(product) = log(left) + log(left+1) +.... log(right)\n        //e^(sum(k = left to right )(ln(k))) -> approx first 5 digits with \n   \n        double logSum = 0;\n        double logSumB10 = 0;\n        double logSumB2 = 0;\n        double logSumB17 = 0;\n        double logSumB3 = 0;\n        for(int i = left; i <= right; i++) {\n            double log = Math.log(i);\n            logSum += log;\n            logSumB10 += log/NATURAL_LOG_10;\n            logSumB2 += log/NATURAL_LOG_2;\n            logSumB17 += log/NATURAL_LOG_17;\n            logSumB3 += log/NATURAL_LOG_3;\n        }\n        logSumB10 = (logSumB10 + logSum/NATURAL_LOG_10)/2;\n        logSumB2 = (logSumB2 + logSum/NATURAL_LOG_2)/2;\n        logSumB17 = (logSumB17 + logSum/NATURAL_LOG_17)/2;\n        logSumB3 = (logSumB3 + logSum/NATURAL_LOG_3)/2;\n        int twoFactors = 0;\n        int fiveFactors = 0;\n\n        for(int i = left; i <= right; i++) {\n            int curr = i;\n            while(curr % 2 == 0) {\n                curr /= 2;\n                twoFactors++;\n            } \n            while(curr % 5 == 0) {\n                curr /= 5;\n                fiveFactors++;\n            }\n        }\n\n        //each 10 needs 1 five and 1 two factor, so the amount of 10s = min(fiveFactors, twoFactors) \n        \n        int trailingZeros = Math.min(fiveFactors, twoFactors);\n        //for each trailiing zero, string number loses 1 place in length to abbrev, this is equivalent to the e^(logSum)/(10*trailingZeros) = e^(logSum - trailingZeros*ln(10)) -> so do logSum -= trailingZeros*ln(10)\n        double log2Of10 = NATURAL_LOG_10/NATURAL_LOG_2;\n        double log17of10 = NATURAL_LOG_10/NATURAL_LOG_17;\n        double log3of10 = NATURAL_LOG_10/NATURAL_LOG_3;\n\n        logSumB10 -= trailingZeros;\n        logSum -= NATURAL_LOG_10*trailingZeros;\n        logSumB2 -= trailingZeros*log2Of10;\n        logSumB17 -= trailingZeros*log17of10;\n\n        int fivesToIgnore = trailingZeros;\n        int twosToIgnore = trailingZeros;\n\n        //this means that the total product is < max for long values\n        if(logSum <= 43.668) {\n            long product = 1;\n            for(long i = left; i <= right; i++) {\n                long curr = i;\n                while(curr % 5 == 0 && fivesToIgnore > 0) {\n                    curr /= 5;\n                    fivesToIgnore--;\n                }\n                while(curr % 2 == 0 && twosToIgnore > 0) {\n                    curr /= 2;\n                    twosToIgnore--;\n                }\n                product *= curr;\n            }\n\n            String num = String.valueOf(product);\n            \n            return (num.length() > 10 ? num.substring(0, 5) + "..." + num.substring(num.length()-5) : num) + "e" + trailingZeros;\n        }\n\n        \n\n        /*e^logSum = product\n        divide product by 10 until product is 5 digits, those are highest 5 digits\n        e^(logSum)/(10^n) = 5 digits of product and following decimals\n        10^n = e^(n*ln(10)) -> e^(logSum)/(10^n) = e^(logSum)/e^(n*ln(10)) = e^(logSum-n*ln(10))\n        n is the integer that forces five digits.\n        e^x is 5 digits when 10000 <= e^x < 100000 -> 10^4 <= e^x < 10^5 -> 4*ln(10) <= x < 5*ln(10)\n        e^(logSum-n*ln(10)) -> where 4*ln(10) <= logSum-n*ln(10) < 5*ln(10) -> since only subtracting integer multiples of ln(10), exp should stay the same % ln(10)\n        e^(logSum - n*ln(10)) = e^((logSum % ln(10)) + 4*ln(10))\n\n        */\n        double expTest = Math.exp((logSum % NATURAL_LOG_10) + 4.0*NATURAL_LOG_10);\n        double expTestB10 = Math.pow(10, (logSumB10 % 1) + 4);\n        double expTestB2 = Math.pow(2, (logSumB2 % log2Of10) + 4.0 * log2Of10);\n        double expTestB17 = Math.pow(17, (logSumB17 % log17of10) + 4.0 * log17of10);\n        double expTestB3 = Math.pow(3, (logSumB3 % log3of10) + 4.0*log3of10);\n\n        int freq = 0;\n        int val = Math.max(Math.max(Math.max((int)expTest, (int)expTestB10), Math.max((int)expTestB2, (int)expTestB17)), (int)expTestB3);\n        if((int)expTest == val) freq++;\n        if((int)expTestB10 == val) freq++;\n        if((int)expTestB2 == val) freq++;\n        if((int)expTestB17 == val) freq++;\n        if((int)expTestB3 == val) freq++;\n        freq %= 5; //if all of them are max, none surpass to next integer barrier, so freq is equivalent to 0\n\n        //take average of both to make lessen the error of either base\n        double expAVG = (expTest + expTestB10 + expTestB2 + expTestB17 + expTestB3)/5.0;\n\n        String firstFiveDigits = String.valueOf((int)(ERR*freq + expAVG));\n        \n        long prodTest = 1l;\n        \n\n        long modProduct = 1l;\n        long MOD = 100000l;\n\n        for(int i = left; i <= right; i++) {\n            int curr = i;\n            while(curr % 5 == 0 && fivesToIgnore > 0) {\n                curr /= 5;\n                fivesToIgnore--;\n            }\n            while(curr % 2 == 0 && twosToIgnore > 0) {\n                curr /= 2;\n                twosToIgnore--;\n            }\n            modProduct = (modProduct * curr) % MOD;\n            prodTest *= curr; \n            while(prodTest > MOD *10 * i) {\n                prodTest /= 10;\n            }\n        }\n        /*\n        System.out.println(prodTest);\n        System.out.println(expTest);\n        System.out.println(expTestB10);\n        System.out.println(expTestB2);\n         System.out.println(expTestB17);\n         System.out.println(expTestB3);\n        System.out.println(expAVG);\n\n*/\n        String lastFiveDigits = String.valueOf(modProduct);\n        while(lastFiveDigits.length() < 5) {//if leading zeros were cut off in string conversion, add them back\n            lastFiveDigits = "0" + lastFiveDigits;\n        }\n        return firstFiveDigits + "..." + lastFiveDigits + "e" + trailingZeros;\n\n        \n        \n        \n    }\n}\n```
['Java']