W A I T + A L L --------- G I F T S
equals:
9 6 0 8 + 6 7 7 --------- 1 0 2 8 5
Given strings a,b,c find where a + b = c; No left hand zeros. All solutions are believed to be unique.
Beware, the test machine might time out your entry!
If you want some easier problems that build up to this one,
Why are the above building blocks to solving this problem? Well, let's think about the simplest thing that could possibly work.
If we
Eventually, we would find the correct answer. Worst case scenario, we have a one in 10! (1/3,600,000) chance of stumbling upon the answer. For the eight character case, it is 8! (1/40,320). I like those odds enough that it is worth trying.
A supplemental problem finds out which technique, methodical or random is better.
The problem is to find the number-to-string matching that results in a + b = 3, right?
Test case needs to be changed,thre can be more than 1 solution to the problem.
e.g. in question for example,[afgilstw = 74062195] is also one correct solution.
Suresh, Which test case? Please be more explicit. These were all taken from a list of known problems, so I need more info to understand if one is flawed.
I am saying in general there can be more than one solution e.g. the one in the question, Wait+All = GIFTS , for that[afgilstw = 74062195] is also valid solution. i.e.[ 5769 +722 = 6491]
Note in the instructions: "No left hand zeros"
Difficult to solve in time.
I saw the note regarding "No left hand zeros", but did not put it together until later that it's really a hint regarding the sum; the 'G' in gifts cannot be mapped to a 0. It's clear now, but initially I thought the remark was referring to 'ALL' not having a blank character to the left (e.g. '\ ALL') that would be 0 when mapped to digits (e.g. 0677).
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Replace all zeros and NaNs in a matrix with the string 'error'
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