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    <p>Here's a simple algorithm which scans the array once, constantly checking if its currently-seen covering range is shorter than the previously seen covering ranges.</p> <p>For simplicity, I'm going to assume that we can map A, B, C, D, and E to the integers 0-4 so that we can easily reference an array. I haven't checked it thoroughly, so please mentally/actually run it on an example or two to make sure it does what you want.</p> <pre><code>//Cell 0 is the last index at which we saw an A, cell 1 " " saw a B, etc. int[] mostRecent = new int[U.length]; mostRecent.setAllValsTo(POSITIVE_INFINITY); int shortestRange = POSITIVE_INFINITY; //We are trying to minimize this number. int minIndex = 0; //The beginning index of the range int maxIndex = POSITIVE_INFINITY; //The ending index of the range. for(int i=0; i&lt; S.length; i++) { int currentValue = S[i]; //This value will be 0-4, corresponding to A-E mostRecent[currentValue] = i; currentMax = mostRecent.findMax(); //beginning of current range currentMin = mostRecent.findMin(); //end of current range currentRange = currentMax - currentMin; if(currentRange &lt; shortestRange) { shortestRange = currentRage; minIndex = currentMin; maxIndex = currentMax; } } //currentMax and currentMin now carry the starting and ending indices, use them as you see fit. return shortestRange; </code></pre> <p>This is order O(nk), with n=S.length and k=U.length. There are still plenty of optimizations to squeeze out of it, but I don't know if I could bring the worst-case order lower.</p>
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    1. POFind the shortest path in one dimension
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    1. COThough This Algorithm Finds The Shortest DISTANCE of the path ,It Doesn't Find the PATH itself
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    2. COOh, that's easy though. Just remember the min and max indices as you go along. Here, I'll modify the code.
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